I will begin by an giving some motivation:
Let $(x,y)$ be some point in the plane, the polar coordinates are defined implicitly by $$r \cos \theta=x, \\r \sin \theta =y.$$ Suppose $r \neq 0$, and define $X=x/r,Y=y/r$. We then have $$\cos \theta=X, \\ \sin \theta =Y, $$ and $X^2+Y^2=1$. Taking the differential of both equations gives $$-\sin \theta \mathrm{d} \theta=\mathrm{d} X, \\ \cos(\theta) \mathrm{d} \theta=dY.$$ It is now evident that $$\mathrm{d} \theta=-Y \mathrm{d} X+X \mathrm{d} Y ,$$ and this is a derivation of that well-known result. An interesting point here is that $\mathrm{d} \theta$ is well defined, even though $\theta$ itself isn't a single-valued function.
Now to my analogous question: Suppose instead that $X,Y$ lie on the Weierstrass elliptic curve $$Y^2=4X^3-g_2 X-g_3 $$ for some constants $g_2,g_3$. Then, there is some $\theta$, depending on $X$ and $Y$ satisfying $$\wp(\theta;g_2,g_3)=X, \\ \wp'(\theta;g_2,g_3)=Y, $$ with $\wp$ being the Weierstrass P-function. Taking the differentials, we get $$\wp'(\theta;g_2,g_3) \mathrm{d} \theta=\mathrm{d} X, \\ \wp''(\theta;g_2,g_3) \mathrm{d} \theta=\mathrm{d} Y.$$
Furthermore, using the 2nd order ODE for $\wp$, the pair of equations simplifies to $$\wp'(\theta;g_2,g_3) \mathrm{d} \theta=\mathrm{d} X, \\ \left( 6 \wp^2(\theta;g_2,g_3)-\frac{1}{2} g_2 \right) \mathrm{d} \theta=\mathrm{d}Y $$ I tried to manipulate these two algebraically to get a "nice" expression for $\mathrm{d} \theta$ in this case, but I couldn't do it. I was wondering if there's any simple expression for $\mathrm{d} \theta$ after all, and if there is, I'd like to see it. Thank you!
Is $\quad d\theta = (-3Y dX + 2 X dY)/( 2g_2 X+3g_3)\quad$ nice enough?