= Solution
First make the invertible linear change of <homogeneous coordinates>
$$
U=-y,\qquad V=\sqrt3\,z,\qquad W=x.
$$
The original equation becomes
$$
V^2W=U^3-W^3=(U-W)(U-\omega W)(U-\omega^2W),
\qquad \omega=\exp(2\pi\mathrm i/3).
$$
This takes the <flex> $[0:0:1]$ to the point at infinity $[0:1:0]$ of a <Weierstrass model>.
Put $h=\omega-1$ and choose $s\in\mathbb C$ with $s^2=h^3$. Both are nonzero. Define
$$
X=\frac{U-W}{h},\qquad Y=\frac V{s},\qquad Z=W.
$$
Since $U=hX+Z$, the three factors on the right are $hX$, $h(X-Z)$ and $h(X-\lambda Z)$, where
$$
\boxed{\lambda=\frac{\omega^2-1}{\omega-1}=1+\omega=\frac{1+\mathrm i\sqrt3}{2}.}
$$
The resulting equation is precisely
$$
\boxed{Y^2Z-X(X-Z)(X-\lambda Z)=0,}
$$
the <Legendre form of an elliptic curve>. In particular, $\lambda\ne0,1$.
Combining the two changes gives an explicit <projective transformation> from the original coordinates:
$$
\begin{pmatrix}X\\Y\\Z\end{pmatrix}
=\begin{pmatrix}
-1/h&-1/h&0\\
0&0&\sqrt3/s\\
1&0&0
\end{pmatrix}
\begin{pmatrix}x\\y\\z\end{pmatrix}.
$$
Its <determinant> is $-\sqrt3/(hs)\ne0$, so this is a valid transformation of the whole <projective plane>, including the points where an affine coordinate would be undefined.
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