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.