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Note added on 26 Nov 2018: I have corrected my answer, which had a serious mistake.

For simplicity of notation, let $(x,y,z) = (x_1,x_2,x_3)$. The Hessian form associated to $f_0 = {x_1}^3+{x_2}^3+{x_3}^3+6x_1x_2x_3$ is $$ H(f_0) = \frac{\partial^2f_0}{\partial x_i\partial x_j}\,\mathrm{d}x_i\circ\mathrm{d}x_j\,. $$ The determinant of this Hessian form is easily computed to be $$ \Delta = -6^3 ({x_1}^3+{x_2}^3+{x_3}^3-3x_1x_2x_3)\, (\mathrm{d}x_1\wedge\mathrm{d}x_2\wedge\mathrm{d}x_3)^{\otimes2}. $$ Any linear transformation of $\mathbb{C}^3$ that preserves $f_0$ must preserve $\Delta$. Since $$ {x_1}^3+{x_2}^3+{x_3}^3-3x_1x_2x_3 = (x_1{+}x_2{+}x_3)(x_1{+}sx_2{+}s^2x_3)(x_1{+}s^2x_2{+}sx_3) $$ where $s^2+s+1=0$ (so that $s^3=1$), it follows that any linear transformation $L$ of $\mathbb{C}^3$ that preserves $f_0$ must must permute and scale the three linearly independent forms $$ \xi_i = (x_1{+}s^ix_2{+}s^{2i}x_3)\quad i = 0,1,2. $$ Thus, if $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$ for $i=0,1,2$, it follows that $L^*(\Delta) = (\lambda_1\lambda_2\lambda_3)^3\Delta$, so $(\lambda_1\lambda_2\lambda_3)^3=1$ (since the determinant of a permuation matrix is $\pm1$). Let $h = {x_1}^3{+}{x_2}^3{+}{x_3}^3{-}3x_1x_2x_3 =\xi_0\xi_1\xi_2$. Thus, $L^*h = \lambda h$, where $\lambda = \lambda_1\lambda_2\lambda_3$, a cube root of unity. Expressing $f_0$ in terms of the $\xi_i$, we find that $f_0 = \tfrac13({\xi_0}^3{+}{\xi_1}^3{+}{\xi_2}^3)$. Since $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$, it follows that $\lambda_i^3 = 1$ for $i=0,1,2$.

Conversely, for any permuation $\pi$ of $(0,1,2)$ and any cube roots of unity $\lambda_i$ for $i=0,1,2$, the formula $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$ defines a transformation that preserves $f_0$. Thus, the stabilizer of $f_0$ is a finite group of order $27\cdot 6 = 162$ (which is unitary for the Hermitian form $F=|\xi_1|^2{+}|\xi_2|^2{+}|\xi_3|^2 = 3(|x_1|^2{+}|x_2|^2{+}|x_3|^2)$). Note, though, that this shows that the first sentence of GreginGre's answer is incorrect (and renders the rest of that argument moot).

Meanwhile, we find $$ f_1 = \tfrac13({\xi_0}^3+s^2\,{\xi_1}^3+s^4\,{\xi_2}^3) \qquad\text{and}\qquad f_2 = \tfrac13({\xi_0}^3+s\,{\xi_1}^3+s^2\,{\xi_2}^3). $$ It follows that $L$, as defined above, preserves $(f_i)^3$ if and only if $\pi$ is an even permutation of $(0,1,2)$. Thus, $G$ has order $81$.

In order to preserve $f_1$ and $f_2$, the permutation $\pi$ must be the identity. Thus, $H$ has order $27$.

One sees that $\mathbb{C}[x,y,z]^G$ contains $f_1f_2$, so it is not equal to $\mathbb{C}[f_0,(f_1)^3,(f_2)^3]$.

On the other hand $\mathbb{C}[x,y,z]^H$ is the group that preserves each of the ${\xi_i}^3$ (since these can be recovered from $f_0,f_1,f_2$ and vice versa) and this is clearly equal to the polynomials in the ${\xi_i}^3$. Hence $\mathbb{C}[x,y,z]^H=\mathbb{C}[f_0,f_1,f_2]$.

I think that answers all your questions.

Note added on 26 Nov 2018: I have corrected my answer, which had a serious mistake.

For simplicity of notation, let $(x,y,z) = (x_1,x_2,x_3)$. The Hessian form associated to $f_0 = {x_1}^3+{x_2}^3+{x_3}^3+6x_1x_2x_3$ is $$ H(f_0) = \frac{\partial^2f_0}{\partial x_i\partial x_j}\,\mathrm{d}x_i\circ\mathrm{d}x_j\,. $$ The determinant of this Hessian form is easily computed to be $$ \Delta = -6^3 ({x_1}^3+{x_2}^3+{x_3}^3-3x_1x_2x_3)\, (\mathrm{d}x_1\wedge\mathrm{d}x_2\wedge\mathrm{d}x_3)^{\otimes2}. $$ Any linear transformation of $\mathbb{C}^3$ that preserves $f_0$ must preserve $\Delta$. Since $$ {x_1}^3{+}{x_2}^3{+}{x_3}^3{-}3x_1x_2x_3 = (x_1{+}x_2{+}x_3)(x_1{+}sx_2{+}s^2x_3)(x_1{+}s^2x_2{+}sx_3) $$ where $s^2+s+1=0$ (so that $s^3=1$), it follows that any linear transformation $L$ of $\mathbb{C}^3$ that preserves $f_0$ must must permute and scale the three linearly independent forms $$ \xi_i = (x_1{+}s^ix_2{+}s^{2i}x_3)\quad i = 0,1,2. $$ Thus, $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$ for $i=0,1,2$, where the $\lambda_i$ are nonzero and $\pi$ is a permutation of $\{0,1,2\}$. Expressing $f_0$ in terms of the $\xi_i$, we find that $f_0 = \tfrac13({\xi_0}^3{+}{\xi_1}^3{+}{\xi_2}^3)$. It now follows that $\lambda_i^3 = 1$ for $i=0,1,2$.

Conversely, for any permuation $\pi$ of $\{0,1,2\}$ and any cube roots of unity $\lambda_i$ for $i=0,1,2$, the formula $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$ defines a linear transformation of $\mathbb{C}^3$ that preserves $f_0$. Thus, the stabilizer of $f_0$ is a finite group of order $27\cdot 6 = 162$ (which is unitary for the Hermitian form $F=|\xi_1|^2{+}|\xi_2|^2{+}|\xi_3|^2 = 3(|x_1|^2{+}|x_2|^2{+}|x_3|^2)$). Note, though, that this shows that the first sentence of GreginGre's answer is incorrect (and renders the rest of that argument moot).

Meanwhile, we find $$ f_1 = \tfrac13({\xi_0}^3+s^2\,{\xi_1}^3+s^4\,{\xi_2}^3) \qquad\text{and}\qquad f_2 = \tfrac13({\xi_0}^3+s\,{\xi_1}^3+s^2\,{\xi_2}^3). $$ It follows that $L$, as defined above, preserves $(f_i)^3$ if and only if $\pi$ is an even permutation of $\{0,1,2\}$. Thus, $G$ has order $81$.

In order to preserve $f_1$ and $f_2$, the permutation $\pi$ must be the identity. Thus, $H$ has order $27$.

One sees that $\mathbb{C}[x,y,z]^G$ contains $f_1f_2$, so it is not equal to $\mathbb{C}[f_0,(f_1)^3,(f_2)^3]$.

On the other hand $H$ is the group that preserves each of the ${\xi_i}^3$ (since these are linear combinations of $f_0,f_1,f_2$ and vice versa) and $H$ preserves a monomial ${\xi_0}^{n_0}{\xi_1}^{n_1}{\xi_2}^{n_2}$ if and only if all of the $n_i$ are divisible by $3$. Hence $\mathbb{C}[x,y,z]^H=\mathbb{C}[{\xi_0}^3,{\xi_1}^3,{\xi_2}^3]=\mathbb{C}[f_0,f_1,f_2]$.

I think that answers all your questions.

Note added on 26 Nov 2018: I have corrected my answer, which had a serious mistake.

For simplicity of notation, let $(x,y,z) = (x_1,x_2,x_3)$. The Hessian form associated to $f_0 = {x_1}^3+{x_2}^3+{x_3}^3+6x_1x_2x_3$ is $$ H(f_0) = \frac{\partial^2f_0}{\partial x_i\partial x_j}\,\mathrm{d}x_i\circ\mathrm{d}x_j\,. $$ The determinant of this Hessian form is easily computed to be $$ \Delta = -6^3 ({x_1}^3+{x_2}^3+{x_3}^3-3x_1x_2x_3)\, (\mathrm{d}x_1\wedge\mathrm{d}x_2\wedge\mathrm{d}x_3)^{\otimes2}. $$ Any linear transformation of $\mathbb{C}^3$ that preserves $f_0$ must preserve $\Delta$. Since $$ {x_1}^3+{x_2}^3+{x_3}^3-3x_1x_2x_3 = (x_1{+}x_2{+}x_3)(x_1{+}sx_2{+}s^2x_3)(x_1{+}s^2x_2{+}sx_3) $$ where $s^2+s+1=0$ (so that $s^3=1$), it follows that any linear transformation $L$ of $\mathbb{C}^3$ that preserves $f_0$ must must permute and scale the three linearly independent forms $$ \xi_i = (x_1{+}s^ix_2{+}s^{2i}x_3)\quad i = 0,1,2. $$ Thus, if $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$ for $i=0,1,2$, it follows that $L^*(\Delta) = (\lambda_1\lambda_2\lambda_3)^3\Delta$, so $(\lambda_1\lambda_2\lambda_3)^3=1$ (since the determinant of a permuation matrix is $\pm1$). Let $h = {x_1}^3{+}{x_2}^3{+}{x_3}^3{-}3x_1x_2x_3 =\xi_0\xi_1\xi_2$. Thus, $L^*h = \lambda h$, where $\lambda = \lambda_1\lambda_2\lambda_3$, a cube root of unity. Expressing $f_0$ in terms of the $\xi_i$, we find that $f_0 = \tfrac13({\xi_0}^3{+}{\xi_1}^3{+}{\xi_2}^3)$. Since $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$, it follows that $\lambda_i^3 = 1$ for $i=0,1,2$.

Conversely, for any permuation $\pi$ of $(0,1,2)$ and any cube roots of unity $\lambda_i$ for $i=0,1,2$, the formula $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$ defines a transformation that preserves $f_0$. Thus, the stabilizer of $f_0$ is a finite group of order $27\cdot 6 = 162$ (which is unitary for the Hermitian form $H=|\xi_1|^2{+}|\xi_2|^2{+}|\xi_3|^2 = 3(|x_1|^2{+}|x_2|^2{+}|x_3|^2)$). Note, though, that this shows that the first sentence of GreginGre's answer is incorrect (and renders the rest of that argument moot).

Meanwhile, we find $$ f_1 = \tfrac13({\xi_0}^3+s^2\,{\xi_1}^3+s^4\,{\xi_2}^3) \qquad\text{and}\qquad f_2 = \tfrac13({\xi_0}^3+s\,{\xi_1}^3+s^2\,{\xi_2}^3). $$ It follows that $L$, as defined above, preserves $(f_i)^3$ if and only if $\pi$ is an even permutation of $(0,1,2)$. Thus, $G$ has order $81$.

In order to preserve $f_1$ and $f_2$, the permutation $\pi$ must be the identity. Thus, $H$ has order $27$.

One sees that $\mathbb{C}[x,y,z]^G$ contains $f_1f_2$, so it is not equal to $\mathbb{C}[f_0,(f_1)^3,(f_2)^3]$.

On the other hand $\mathbb{C}[x,y,z]^H$ is the group that preserves each of the ${\xi_i}^3$ (since these can be recovered from $f_0,f_1,f_2$ and vice versa) and this is clearly equal to the polynomials in the ${\xi_i}^3$. Hence $\mathbb{C}[x,y,z]^H=\mathbb{C}[f_0,f_1,f_2]$.

I think that answers all your questions.

Note added on 26 Nov 2018: I have corrected my answer, which had a serious mistake.

For simplicity of notation, let $(x,y,z) = (x_1,x_2,x_3)$. The Hessian form associated to $f_0 = {x_1}^3+{x_2}^3+{x_3}^3+6x_1x_2x_3$ is $$ H(f_0) = \frac{\partial^2f_0}{\partial x_i\partial x_j}\,\mathrm{d}x_i\circ\mathrm{d}x_j\,. $$ The determinant of this Hessian form is easily computed to be $$ \Delta = -6^3 ({x_1}^3+{x_2}^3+{x_3}^3-3x_1x_2x_3)\, (\mathrm{d}x_1\wedge\mathrm{d}x_2\wedge\mathrm{d}x_3)^{\otimes2}. $$ Any linear transformation of $\mathbb{C}^3$ that preserves $f_0$ must preserve $\Delta$. Since $$ {x_1}^3+{x_2}^3+{x_3}^3-3x_1x_2x_3 = (x_1{+}x_2{+}x_3)(x_1{+}sx_2{+}s^2x_3)(x_1{+}s^2x_2{+}sx_3) $$ where $s^2+s+1=0$ (so that $s^3=1$), it follows that any linear transformation $L$ of $\mathbb{C}^3$ that preserves $f_0$ must must permute and scale the three linearly independent forms $$ \xi_i = (x_1{+}s^ix_2{+}s^{2i}x_3)\quad i = 0,1,2. $$ Thus, if $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$ for $i=0,1,2$, it follows that $L^*(\Delta) = (\lambda_1\lambda_2\lambda_3)^3\Delta$, so $(\lambda_1\lambda_2\lambda_3)^3=1$ (since the determinant of a permuation matrix is $\pm1$). Let $h = {x_1}^3{+}{x_2}^3{+}{x_3}^3{-}3x_1x_2x_3 =\xi_0\xi_1\xi_2$. Thus, $L^*h = \lambda h$, where $\lambda = \lambda_1\lambda_2\lambda_3$, a cube root of unity. Expressing $f_0$ in terms of the $\xi_i$, we find that $f_0 = \tfrac13({\xi_0}^3{+}{\xi_1}^3{+}{\xi_2}^3)$. Since $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$, it follows that $\lambda_i^3 = 1$ for $i=0,1,2$.

Conversely, for any permuation $\pi$ of $(0,1,2)$ and any cube roots of unity $\lambda_i$ for $i=0,1,2$, the formula $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$ defines a transformation that preserves $f_0$. Thus, the stabilizer of $f_0$ is a finite group of order $27\cdot 6 = 162$ (which is unitary for the Hermitian form $F=|\xi_1|^2{+}|\xi_2|^2{+}|\xi_3|^2 = 3(|x_1|^2{+}|x_2|^2{+}|x_3|^2)$). Note, though, that this shows that the first sentence of GreginGre's answer is incorrect (and renders the rest of that argument moot).

Meanwhile, we find $$ f_1 = \tfrac13({\xi_0}^3+s^2\,{\xi_1}^3+s^4\,{\xi_2}^3) \qquad\text{and}\qquad f_2 = \tfrac13({\xi_0}^3+s\,{\xi_1}^3+s^2\,{\xi_2}^3). $$ It follows that $L$, as defined above, preserves $(f_i)^3$ if and only if $\pi$ is an even permutation of $(0,1,2)$. Thus, $G$ has order $81$.

In order to preserve $f_1$ and $f_2$, the permutation $\pi$ must be the identity. Thus, $H$ has order $27$.

One sees that $\mathbb{C}[x,y,z]^G$ contains $f_1f_2$, so it is not equal to $\mathbb{C}[f_0,(f_1)^3,(f_2)^3]$.

On the other hand $\mathbb{C}[x,y,z]^H$ is the group that preserves each of the ${\xi_i}^3$ (since these can be recovered from $f_0,f_1,f_2$ and vice versa) and this is clearly equal to the polynomials in the ${\xi_i}^3$. Hence $\mathbb{C}[x,y,z]^H=\mathbb{C}[f_0,f_1,f_2]$.

I think that answers all your questions.

Note added on 26 Nov 2018: I have corrected my answer, which had a serious mistake.

For simplicity of notation, let $(x,y,z) = (x_1,x_2,x_3)$. The Hessian form associated to $f_0 = {x_1}^3+{x_2}^3+{x_3}^3+6x_1x_2x_3$ is $$ H(f_0) = \frac{\partial^2f_0}{\partial x_i\partial x_j}\,\mathrm{d}x_i\circ\mathrm{d}x_j\,. $$ The determinant of this Hessian form is easily computed to be $$ \Delta = -6^3 ({x_1}^3+{x_2}^3+{x_3}^3-3x_1x_2x_3)\, (\mathrm{d}x_1\wedge\mathrm{d}x_2\wedge\mathrm{d}x_3)^{\otimes2}. $$ Any linear transformation of $\mathbb{C}^3$ that preserves $f_0$ must preserve $\Delta$. Since $$ {x_1}^3+{x_2}^3+{x_3}^3-3x_1x_2x_3 = (x_1{+}x_2{+}x_3)(x_1{+}sx_2{+}s^2x_3)(x_1{+}s^2x_2{+}sx_3) $$ where $s^2+s+1=0$ (so that $s^3=1$), it follows that any linear transformation $L$ of $\mathbb{C}^3$ that preserves $f_0$ must must permute and scale the three linearly independent forms $$ \xi_i = (x_1{+}s^ix_2{+}s^{2i}x_3)\quad i = 0,1,2. $$ Thus, if $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$ for $i=0,1,2$, it follows that $L^*(\Delta) = (\lambda_1\lambda_2\lambda_3)^3\Delta$, so $(\lambda_1\lambda_2\lambda_3)^3=1$ (since the determinant of a permuation matrix is $\pm1$). Let $h = {x_1}^3{+}{x_2}^3{+}{x_3}^3{-}3x_1x_2x_3 =\xi_0\xi_1\xi_2$. Thus, $L^*h = \lambda h$, where $\lambda = \lambda_1\lambda_2\lambda_3$, a cube root of unity. Expressing $f_0$ in terms of the $\xi_i$, we find that $f_0 = \tfrac13({\xi_0}^3{+}{\xi_1}^3{+}{\xi_2}^3)$. Since $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$, it follows that $\lambda_i^3 = 1$ for $i=0,1,2$.

Conversely, for any permuation $\pi$ of $(0,1,2)$ and any cube roots of unity $\lambda_i$ for $i=0,1,2$, the formula $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$ defines a transformation that preserves $f_0$. Thus, the stabilizer of $f_0$ is a finite group of order $27\cdot 6 = 162$ (and is therefore necessarily unitary for the appropriate Hermitian form on $\mathbb{C}^3$). Note, though, that this shows that the first sentence of GreginGre's answer is incorrect (and renders the rest of that argument moot).

Meanwhile, we find $$ f_1 = \tfrac13({\xi_0}^3+s^2\,{\xi_1}^3+s^4\,{\xi_2}^3) \qquad\text{and}\qquad f_2 = \tfrac13({\xi_0}^3+s\,{\xi_1}^3+s^2\,{\xi_2}^3). $$ It follows that $L$, as defined above, preserves $(f_i)^3$ if and only if $\pi$ is an even permutation of $(0,1,2)$. Thus, $G$ has order $81$.

In order to preserve $f_1$ and $f_2$, the permutation $\pi$ must be the identity. Thus, $H$ has order $27$.

One sees that $\mathbb{C}[x,y,z]^G$ contains $f_1f_2$, so it is not equal to $\mathbb{C}[f_0,(f_1)^3,(f_2)^3]$.

On the other hand $\mathbb{C}[x,y,z]^H$ is the group that preserves each of the ${\xi_i}^3$ (since these can be recovered from $f_0,f_1,f_2$ and vice versa) and this is clearly equal to the polynomials in the ${\xi_i}^3$. Hence $\mathbb{C}[x,y,z]^H=\mathbb{C}[f_0,f_1,f_2]$.

I think that answers all your questions.

Note added on 26 Nov 2018: I have corrected my answer, which had a serious mistake.

For simplicity of notation, let $(x,y,z) = (x_1,x_2,x_3)$. The Hessian form associated to $f_0 = {x_1}^3+{x_2}^3+{x_3}^3+6x_1x_2x_3$ is $$ H(f_0) = \frac{\partial^2f_0}{\partial x_i\partial x_j}\,\mathrm{d}x_i\circ\mathrm{d}x_j\,. $$ The determinant of this Hessian form is easily computed to be $$ \Delta = -6^3 ({x_1}^3+{x_2}^3+{x_3}^3-3x_1x_2x_3)\, (\mathrm{d}x_1\wedge\mathrm{d}x_2\wedge\mathrm{d}x_3)^{\otimes2}. $$ Any linear transformation of $\mathbb{C}^3$ that preserves $f_0$ must preserve $\Delta$. Since $$ {x_1}^3+{x_2}^3+{x_3}^3-3x_1x_2x_3 = (x_1{+}x_2{+}x_3)(x_1{+}sx_2{+}s^2x_3)(x_1{+}s^2x_2{+}sx_3) $$ where $s^2+s+1=0$ (so that $s^3=1$), it follows that any linear transformation $L$ of $\mathbb{C}^3$ that preserves $f_0$ must must permute and scale the three linearly independent forms $$ \xi_i = (x_1{+}s^ix_2{+}s^{2i}x_3)\quad i = 0,1,2. $$ Thus, if $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$ for $i=0,1,2$, it follows that $L^*(\Delta) = (\lambda_1\lambda_2\lambda_3)^3\Delta$, so $(\lambda_1\lambda_2\lambda_3)^3=1$ (since the determinant of a permuation matrix is $\pm1$). Let $h = {x_1}^3{+}{x_2}^3{+}{x_3}^3{-}3x_1x_2x_3 =\xi_0\xi_1\xi_2$. Thus, $L^*h = \lambda h$, where $\lambda = \lambda_1\lambda_2\lambda_3$, a cube root of unity. Expressing $f_0$ in terms of the $\xi_i$, we find that $f_0 = \tfrac13({\xi_0}^3{+}{\xi_1}^3{+}{\xi_2}^3)$. Since $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$, it follows that $\lambda_i^3 = 1$ for $i=0,1,2$.

Conversely, for any permuation $\pi$ of $(0,1,2)$ and any cube roots of unity $\lambda_i$ for $i=0,1,2$, the formula $L^*(\xi_i) = \lambda_i\,\xi_{\pi(i)}$ defines a transformation that preserves $f_0$. Thus, the stabilizer of $f_0$ is a finite group of order $27\cdot 6 = 162$ (which is unitary for the Hermitian form $H=|\xi_1|^2{+}|\xi_2|^2{+}|\xi_3|^2 = 3(|x_1|^2{+}|x_2|^2{+}|x_3|^2)$). Note, though, that this shows that the first sentence of GreginGre's answer is incorrect (and renders the rest of that argument moot).

Meanwhile, we find $$ f_1 = \tfrac13({\xi_0}^3+s^2\,{\xi_1}^3+s^4\,{\xi_2}^3) \qquad\text{and}\qquad f_2 = \tfrac13({\xi_0}^3+s\,{\xi_1}^3+s^2\,{\xi_2}^3). $$ It follows that $L$, as defined above, preserves $(f_i)^3$ if and only if $\pi$ is an even permutation of $(0,1,2)$. Thus, $G$ has order $81$.

In order to preserve $f_1$ and $f_2$, the permutation $\pi$ must be the identity. Thus, $H$ has order $27$.

One sees that $\mathbb{C}[x,y,z]^G$ contains $f_1f_2$, so it is not equal to $\mathbb{C}[f_0,(f_1)^3,(f_2)^3]$.

On the other hand $\mathbb{C}[x,y,z]^H$ is the group that preserves each of the ${\xi_i}^3$ (since these can be recovered from $f_0,f_1,f_2$ and vice versa) and this is clearly equal to the polynomials in the ${\xi_i}^3$. Hence $\mathbb{C}[x,y,z]^H=\mathbb{C}[f_0,f_1,f_2]$.

I think that answers all your questions.