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 $(\lambda_1\lambda_2\lambda_3)^3=1$.  (After all, the determinant of a permuation matrix is $\pm1$.)  Setting $h = {x_1}^3+{x_2}^3+{x_3}^3-3x_1x_2x_3 =\xi_0\xi_1\xi_2$, it follows that $L^*h = \lambda h$, where $\lambda = \lambda_1\lambda_2\lambda_3$, a cube root of unity.

Now, $f_0-h = 9x_1x_2x_3$ , but it is easy to see that the only linear combinations of $h$ and $f_0$ that factor into linear factors are $h$ and $f_0-h$.  Now, if a linear transformation $L$ of $\mathbb{C}$ satisfies $L^*f_0 = f_0$ and $L^*h = \lambda h$, then $L^*(f_0-h) = f_0-\lambda h$ must factor into linear factors, so it follows that $\lambda = 1$.  Consequently $L^*(x_1x_2x_3) = x_1x_2x_3$, so $L$ must permute and scale the variables $x_i$.  Meanwhile, $f_0+2h = 3({x_1}^3+{x_2}^3+{x_3}^3)$, so it follows that the numbers $\mu_i$ such
that $L^*x_i = \mu_i x_{\pi(i)}$ must satisfy $\mu_i^3 = 1$ and $\mu_1\mu_2\mu_3=1$.  

This proves the first sentence of GreginGre's answer.  However, note that the second sentence is false, since, if $L^*(x,y,z) = (x,z,y)$, then $L^*f_1 = f_2$
and $L^*f_2 = f_1$.  In fact, if we want $L^*f_0=f_0$ and $L^*f_1 = \nu f_1$
for some complex number $\nu$, then $L^*x_i = \mu_i x_{\pi(i)}$ must have $\pi$ be an *even* permutation of $[1,2,3]$.  If $L^*(x_1,x_2,x_3) = (\mu_1x_2,\mu_2x_3,\mu_3x_1)$, then we must have $\nu = \mu_2/\mu_1 = \mu_3/\mu_2=\mu_1/\mu_3$, so $\mu_2 = \nu\mu_1$, $\mu_3 =\nu^2\mu_2$, and, of course, $\mu_1 = \nu^3\mu_1$, so $\nu^3 = 1$.  Thus, we must have
$$
L^*(x_1,x_2,x_3) = (\mu_1x_2,\nu\mu_1x_3,\nu^2\mu_1x_1).
$$
Note that $(L^2)^*f_1 = \nu^2 f_1$.
Of course, this $L$ will also satisfy $L^*f_2 = \nu^2f_2$ and $(L^2)^*f_2 = \nu f_2$.

It easily follows now that $G$ has order $27$ while $H$ has order $9$.  It's clear that every element of $G$ is unitary.

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

It is also not true that $\mathbb{C}[x,y,z]^H$ is equal to $\mathbb{C}[f_0,f_1,f_2]$, since it must also contain $f_0-h = 9xyz$.

I think that answers all your questions.