= Solution
For $F=x^3+y^3+3xz^2$, the first <partial derivatives> are
$$
F_x=3(x^2+z^2),\qquad F_y=3y^2,\qquad F_z=6xz.
$$
Their simultaneous vanishing would force $y=0$, $xz=0$ and $x^2+z^2=0$, hence $x=y=z=0$, which is not a <projective point>. This also verifies that the given <plane cubic> is nonsingular.
The <Hessian matrix> and its <determinant> are
$$
\operatorname{Hess}F=6\begin{pmatrix}x&0&z\\0&y&0\\z&0&x\end{pmatrix},
\qquad H_F=216y(x^2-z^2).
$$
Thus the <Hessian curve> is the union of $y=0$, $x=z$ and $x=-z$. On $y=0$, the equation of $C$ reduces to $x(x^2+3z^2)=0$, giving
$$
\boxed{[0:0:1],\quad[\mathrm i\sqrt3:0:1],\quad[-\mathrm i\sqrt3:0:1].}
$$
On $x=z$ one must have $z\ne0$, so put $z=x=1$ and obtain $y^3=-4$. On $x=-z$ similarly put $z=1$, $x=-1$, obtaining $y^3=4$. Choose $r\in\mathbb C$ with $r^3=4$ and let $\omega=\exp(2\pi\mathrm i/3)$. The remaining points are
$$
\boxed{[1:-r\omega^j:1],\quad[-1:r\omega^j:1],\qquad j=0,1,2.}
$$
All nine listed <projective points> are distinct. They exhaust the intersection with the <Hessian curve>, so the <Hessian criterion for a flex> proves that these, and only these, are the <flexes> in this example. There are none at infinity: $z=0$ and $H_F=0$ would force $x=y=0$ on $C$.
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