Differentiating the inverse Weierstrass P-function I will begin with some background:
The solutions $\theta$ of $$\cos \theta=x $$
constitute of two families, each of which is an arithmetic progression. Namely, if $\arccos x$ denotes any particular solution then the families are
$$\left\{ \arccos x+2 \pi k: k \in \mathbb{Z} \right\} \cup \left\{ -\arccos x+2 \pi k: k \in \mathbb{Z} \right\}. $$
If, instead, we solve the system
$$\cos \theta=x \\ \sin \theta=y $$
with $x,y$ satisfying the compatibility condition $$x^2+y^2=1 $$ then the solutions $\theta$ form a single arithmetic progression
$$\{\theta^*+2 \pi k: k \in \mathbb{Z} \} $$
where $\theta^*$ is any particular solution. Since $\theta$ is then uniquely defined up to additive integer multiples of $2 \pi$, the differential $\mathrm{d} \theta$ is well-defined, and one can see that
$$\mathrm{d} \theta=-y \mathrm{d} x+x \mathrm{d} y. $$

Analogously, one can think of the equation $$\wp(\theta;g_2,g_3)=x $$
as being an underdetermined system, since it has the reflection symmetry $\theta \mapsto -\theta$, as well as the translations $\theta \mapsto \theta+2 \omega_i, \; i=1,2$, where $2\omega_1,2\omega_2$ are generators of the period lattice $$\Lambda=\{2m \omega_1+2n \omega_2:m,n \in \mathbb{Z} \}.$$
If, instead one looks at the system
$$\wp(\theta;g_2,g_3)=x \\ \wp'(\theta;g_2,g_3)=y$$
with $x,y$ satisfying the compatibility condition $$y^2=4x^3-g_2 x-g_3 $$
then $\theta$ is uniquely defined only up to translations by points on the period lattice. In fact, one of the branches of $\theta$ is what Mathematica calls 
InverseWeierstrassP[{x,y},{g2,g3}] or $$\wp^{-1} (x,y;g_2,g_3).$$
My questions are:


*

*Is there a nice formula for $\mathrm{d} \theta$ in this case?

*Similarly, are there simple formulas for the partial derivatives below? $$\frac{\partial}{\partial x} \wp^{-1}(x,y;g_2,4x^3-g_2x-y^2) \\ \frac{\partial}{\partial y} \wp^{-1}(x,y;g_2,4x^3-g_2x-y^2) $$

 A: Edit and correction (thanks to Nemo for pointing out):
First note that even though $d\theta$ is well defined, it is well defined as a differential of a map on $S^1$ and not on $\mathbb{R}^2$, which means that its representation as a linear sum of $dx$ and $dy$ is not unique. 
In fact, the tangent space (and the cotangent space) are one-dimensional.
Therefore, the choice of map varies the differential (as a differential on $\mathbb{R}^2$) - e.g. the linear combination you have written can be obtained by using $\theta = \arctan(y/x)$.
In our case, the explicit description of this branch of the inverse Weierstrass function is $\wp^{-1}(x,y)=\int_{-\infty}^{x} \frac{1}{\sqrt{4t^3-g_2t-g_3}}dt$.
This means that
$$ d\theta = y^{-1}dx + (6x^2 - \frac{g_2}{2})^{-1}dy $$
This makes sense regarding the previous answer, where I had forgotten to invert the results in the end.
Old Answer:
Note that the differential $d\theta$ is simply the gradient of the path defined above. In the case of $\wp$ we see that the path is given by $\theta \mapsto (\wp(\theta), \wp'(\theta)) = (x,y)$, hence the gradient is given by $(\wp'(\theta), \wp''(\theta))$.
Next, we derive the functional equation
$$ (\wp'(\theta))^2 = 4(\wp(\theta))^3 - g_2 \wp(\theta) - g_3 $$
to obtain
$$ 2 \wp'(\theta) \wp''(\theta) = 12 (\wp(\theta))^2 \wp'(\theta) - g_2 \wp'(\theta) $$
and hence
$$ \wp''(\theta) = 6(\wp(\theta))^2 - \frac{g_2}{2} $$
Recalling that $\wp(\theta) = x$, we see that $\wp''(\theta) = 6x^2-\frac{g_2}{2}$, hence we may write
$$ d\theta = ydx + (6x^2 - \frac{g_2}{2})dy $$
Hope that this fully answers your question.
