second order partial derivitives The example I am using is polar coordinates, however my confusion can be extended.
so we have
$f(x,y)$ where $x = x(r,\theta)$ and $y = y(r,\theta) $
The first derivative I understand:
$$\left(\frac{\partial f}{\partial x}\right)_y=\left(\frac{\partial f}{\partial r}\right)_\theta\left(\frac{\partial r}{\partial x}\right)_y+\left(\frac{\partial f}{\partial \theta}\right)_r\left(\frac{\partial \theta}{\partial x}\right)_y$$
now if we differentiate with respect to x again - I have used the product rule on both functions in each term:
$$\left(\frac{\partial^2 f}{\partial x^2}\right)_y=
\left(\frac{\partial^2 f}{\partial r^2}\right)_\theta\left(\frac{\partial r}{\partial x}\right)_y^2 +
\left(\frac{\partial f}{\partial r}\right)_\theta\left(\frac{\partial^2 r}{\partial x^2}\right)_y+
\left(\frac{\partial^2 f}{\partial \theta^2}\right)_r\left(\frac{\partial \theta}{\partial x}\right)_y^2 +
\left(\frac{\partial f}{\partial \theta}\right)_r\left(\frac{\partial^2 \theta}{\partial x^2}\right)_y
+$$
$$
+2\left(\frac{\partial^2 f}{\partial r \partial \theta}\right)_{\theta r}\,\left(\frac{\partial r}{\partial x}\right)_y \,\left(\frac{\partial \theta}{\partial x}\right)_y 
$$
However, I believe this to be wrong as we can take 
$$\left(\frac{\partial }{\partial x}\right)_y=\left(\frac{\partial }{\partial r}\right)_\theta\left(\frac{\partial r}{\partial x}\right)_y+\left(\frac{\partial }{\partial \theta}\right)_r\left(\frac{\partial \theta}{\partial x}\right)_y$$
Applying this to $(\frac{\partial f}{\partial x})_y$ we get:
$$(\frac{\partial^2 f}{\partial x^2})_y=
(\frac{\partial^2 f}{\partial r^2})_\theta(\frac{\partial r}{\partial x})_y^2 +
(\frac{\partial^2 f}{\partial \theta^2})_r(\frac{\partial \theta}{\partial x})_y^2 +
2(\frac{\partial^2 f}{\partial r \partial \theta})_{\theta r}(\frac{\partial r}{\partial x})_y (\frac{\partial \theta}{\partial x})_y 
$$
Where did the
$$
(\frac{\partial f}{\partial r})_\theta(\frac{\partial^2 r}{\partial x^2})_y\;\;,\;\;\;\;
(\frac{\partial f}{\partial \theta})_r(\frac{\partial^2 \theta}{\partial x^2})_y$$
terms go?  Its as if the product rule doesn't hold, the $\;
(\frac{\partial r}{\partial x})_y\;\;,\;\;\;
(\frac{\partial \theta}{\partial x})_y\;$  didn't get differentiated.
Where is my mistake?  thank you

First, I think we have:
$$f'_x=f'_r\cdot r'_x+f'_\theta\cdot\theta'_x=f'_r\cos\theta-f'_\theta\frac{\sin\theta}r$$
so your first (second order mixed) derivative is:
$$f'_{xy}=\left(f'_r\right)'_y\,r'_x+f'_r\,(r'_x)'_y+(f'_\theta)'_y\,\theta'_x+f'_\theta\,(\theta_x)'_y$$
For example:
$$\begin{cases}x=r\cos\theta\\y=r\sin\theta\end{cases}\implies r=\sqrt{x^2+y^2}\implies r'_x=\frac x{\sqrt{x^2+y^2}}=\sin\theta\;,\;\;$$
$$(r'_x)'_y=-\frac{xy}{(x^2+y^2)^{3/2}}=-\frac{\cos\theta\sin\theta}r=-\frac1{2r}\,\sin2\theta$$
and etc.
