Simple partial differentiation $x = r\cos\theta$ and $y = r\sin\theta$ If 
\begin{align}
x &= r\cos\theta,\\
y &= r\sin\theta,
\end{align}
find 
$$\dfrac{\partial^2\theta}{\partial{x}\partial{y}}.$$
How can I find this partial derivative?
I need to prove that 
$$
\frac{\partial^2\theta}{\partial{x}\partial{y}} = -\frac{\cos2\theta}{r^2}.$$

Using $\theta = \arcsin\left(\frac{y}{r}\right)$ and $\arcsin'(t) = \frac{1}{\sqrt{1-t^2}}$ we get
$$ \frac{\partial \theta}{\partial y} = \frac{1}{r} \frac{1}{\sqrt{1- \left(\frac{y}{r}\right)^2}} =  \frac{1}{x}$$
using $r^2 = x^2 + y^2$. Hence
$$\frac{ \partial \theta}{\partial x\partial y} = -\frac{1}{x^2} = -\frac{\cos^2(\theta)}{r^2} $$
