Help with multivar. chain rule I am having trouble with the following problem. I feel that I do understand the multivariable chain rule in general, but applying it here is more difficult. I am lost on where to start. Any help would be appreciated.

Let $f(x,y)$ be a twice continously differentiable function, and let $x=r\cos\theta$, $y=r\sin\theta$.  Show that
$$\frac{\partial^2 f}{\partial x^2}
+\frac{\partial^2 f}{\partial y^2}
=\frac{\partial^2 f}{\partial r^2}
+\frac{1}{r^2} \frac{\partial^2 f}{\partial \theta^2}
+\frac{1}{r} 
\frac{\partial f}{\partial r}$$

P.S. I apologize if my formatting is poor. This is my first time on this site.
 A: Here we actually compute the Right hand side and show that it is equivalent to the left.
Firstly two variable chain rule: $\frac{\partial}{\partial t}(h(g(t,s),f(t,s)))= \frac{\partial h}{\partial g}\frac{\partial g}{\partial t}+\frac{\partial h}{\partial f}\frac{\partial f}{\partial t}$
Now in our context we have $f(x(r,\theta),y(r,\theta))$, so:
$\frac{\partial f}{\partial r}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}=\frac{\partial f}{\partial x}\cos(\theta)+\frac{\partial f}{\partial y}\sin(\theta)$
$\frac{\partial^2 f}{\partial r^2}=\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial r}\cos(\theta)+\frac{\partial^2 f}{\partial y^2}\frac{\partial y}{\partial r}\sin(\theta)=\frac{\partial^2 f}{\partial x^2}\cos^2(\theta)+\frac{\partial^2 f}{\partial y^2}\sin^2(\theta)$
$\frac{\partial f}{\partial \theta}=\frac{\partial f}{\partial x}(-r\sin\theta)+\frac{\partial f}{\partial y}(r\cos\theta)$
$\frac{\partial^2 f}{\partial \theta^2} = -\frac{\partial f}{\partial x}r\cos(\theta)-\frac{\partial f}{\partial y}r\sin(\theta)+r^2\cos^2\theta\frac{\partial^2 f}{\partial x^2}+r^2\sin^2\theta\frac{\partial^2 f}{\partial y^2}$
So:
RHS = $\frac{\partial^2 f}{\partial x^2}\cos^2(\theta)+\frac{\partial^2 f}{\partial y^2}\sin^2(\theta)+\frac{1}{r^2}(-\frac{\partial f}{\partial x}r\cos(\theta)-\frac{\partial f}{\partial y}r\sin(\theta)+r^2\sin^2\theta\frac{\partial^2 f}{\partial x^2}+r^2\cos^2\theta\frac{\partial^2 f}{\partial y^2})$
$+\frac{1}{r}(\frac{\partial f}{\partial x}\cos(\theta)+\frac{\partial f}{\partial y}\sin(\theta))=\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial x^2}$
