Express partial derivatives of second order (and the Laplacian) in polar coordinates [duplicate]

$z=f(x,y)$ where $x=rcosθ$ and $y=rsinθ$

Find $ \frac{\partial z}{\partial x}$ and $ \frac{\partial^2 z}{\partial x^2}$

I'm having big troubles with using chain rule, in particularly the second derivative. Spent 2 hours on this already. What I have...

$ \frac{\partial z}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial z}{\partial r}+\frac{\partial z}{\partial θ}\frac{\partial θ}{\partial x}$. So I'm guessing that $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial θ}$ are unkown? And I will need to use them to differentiate again. Any help is really welcome. This is simply only part of a bigger question.

Full Question:

One source of confusion in calculations like this is that $z$ is being used to denote two different (but related) functions. Let's define a function $\hat{z}$ by \begin{equation} \hat{z}(r,\theta) = z(r \cos \theta, r \sin \theta). \end{equation} Note that $\hat{z}$ is not the same function as $z$, so it should have its own name.

Now it is more clear how to apply the chain rule. I also think the calculation is clarified by writing inputs to functions explicitly. \begin{align} \frac{\partial \hat{z}(r,\theta)}{\partial r} &= \frac{\partial z(r \cos \theta, r \sin \theta)}{\partial x} \cos \theta + \frac{\partial z(r \cos \theta, r \sin \theta)}{\partial y} \sin \theta.\\ \frac{\partial \hat{z}(r,\theta)}{\partial \theta} &= \frac{\partial z(r \cos \theta, r \sin \theta)}{\partial x} (-r \sin \theta) + \frac{\partial z(r \cos \theta, r \sin \theta)}{\partial y} (r \cos \theta). \end{align} Higher order partial derivatives can be computed next. Then we can combine and simplify to get an expression for \begin{equation} \frac{\partial^2 z(r \cos \theta, r \sin \theta)}{\partial x^2} + \frac{\partial^2 z(r \cos \theta, r \sin \theta)}{\partial y^2} \end{equation} in terms of partial derivatives of $\hat{z}$.