Partial derivative chain rules. $$x= \cosθ -r\sinθ$$
$$y= \sinθ + r\cosθ $$
Show that,
$$ \frac{\partial^2θ }{\partial x^2}= \frac{\cosθ }{ r^3} (\cosθ -2r\sinθ)$$
Please Help :)
I used the chain rule: 
$$ \frac{\partialθ }{\partial x}=-\frac{\cosθ }{ r}$$
But now I don't know how to find the second derivative.

First you find $x_r,x_\theta, y_r,y_\theta$. Then you use (inverse function theorem, kind of chain rule) $$\left(\begin{array}{ll}r_x&r_y\\\theta_x&\theta_y\end{array}\right)\left(\begin{array}{ll}x_r&x_\theta\\y_r&y_\theta\end{array}\right)=\left(\begin{array}{ll}1&0\\0&1\end{array}\right)$$
to get $\theta_x=\frac{-y_r}{r}=-\frac{\cos\theta}{r}, r_x=\frac{y_\theta}{r}=\frac{\cos\theta-r\sin\theta}{r}$. Then you apply quotient rule to $\theta_x$ to get $$\theta_{xx}=\frac{r(\sin\theta)\theta_x+r_x\cos\theta}{r^2}$$
Substituting $\theta_x$ and $r_x$ to get the answer.
