**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.