Degree 2 map of elliptic Curve

Assuming $\operatorname{char}k\neq 2,3$.

If we have a cubic in, for example, Weierstrass form $$ Y^2Z+a_1XYZ+a_3YZ^2=X^3+a_2X^2Z+a_4XY^2+a_6Z^3 $$ then it is obvious $(X:Y:Z)\mapsto(X:Z)$ is a degree $2$ map to $\mathbb{P}^1$. So let's try to transform our equation to something like that.

First, move a point on our curve, say $(0:1:1)$, to the point $(0:1:0)$ that you would expect for an elliptic curve in Weierstrass form, so \begin{align*} 0&=X^3+Y^3-Z^3\\ &=X^3+(Y-Z)[Y^2+YZ+Z^2]\\ &=X^3+(Y-Z)\left[\frac14(Y-Z)^2+\frac34(Y+Z)^2\right] \end{align*} Hence we have $$ 3(Y-Z)(Y+Z)^2=-4X^3-(Y-Z)^3 $$ So the map $$ (X:Y:Z)\mapsto(X:Y-Z) $$ is a degree 2 map $C\to\mathbb{P}^1$.