Suppose $u(x,y)$ is a function $C^2$ from $\mathbb{R}^2$ to $\mathbb{R}$. After a change to polar coordinates $u(x,y)=u(r\cos \theta , r\sin \theta)$. Suppose $u(x,y)$ is a function $C^2$ from $\mathbb{R}^2$ to $\mathbb{R}$. After a change to polar coordinates $u(x,y)=u(r\cos \theta , r\sin \theta)$. We have
\begin{align*}
u_r=\frac{\partial u}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial r};\ u_\theta = \frac{\partial u}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial \theta}
\end{align*}
Use the chain rule to show that
\begin{align*}
u_{xx}+u_{yy}=u_{rr}+\frac{1}{r}u_\theta
\end{align*}
Attempt:
\begin{align*}
x&=r\cos \theta \\
y&=r\sin \theta
\end{align*}
\begin{align*}
\frac{\partial x}{\partial r}=\cos \theta && \frac{\partial x}{\partial \theta}=-r\sin \theta \\
\frac{\partial y}{\partial r}=\sin \theta && \frac{\partial y}{\partial \theta}=r\cos \theta
\end{align*}
\begin{align*}
u_r&=\frac{\partial u}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial r} \\
&=\frac{\partial u}{\partial x} \cos \theta + \frac{\partial u}{\partial y}\sin \theta \\
&=\cos \theta\frac{\partial u}{\partial x} + \sin \theta \frac{\partial u}{\partial y}
\end{align*}
\begin{align*}
u_{rr}&=\cos \theta \frac{\partial }{\partial r}\frac{\partial u}{\partial x} + \sin \theta \frac{\partial }{\partial r}\frac{\partial u}{\partial y} \\
&= \cos \theta \left( \frac{\partial }{\partial x}\frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial }{\partial y}\frac{\partial u}{\partial x}\frac{\partial y}{\partial r} \right) + \sin \theta \left( \frac{\partial }{\partial x}\frac{\partial u}{\partial y}\frac{\partial x}{\partial r}+\frac{\partial }{\partial y}\frac{\partial u}{\partial y}\frac{\partial y}{\partial r} \right) \\
&=u_{xx}\cos ^2 \theta +u_{xy}2\cos \theta \sin \theta +u_{yy}\sin ^2 \theta 
\end{align*}
\begin{align*}
u_\theta &= \frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta} \\
&= \frac{\partial u}{\partial x}(-r\sin \theta)+\frac{\partial u}{\partial y}(r\cos \theta) \\
&=r\cos \theta \frac{\partial u}{\partial y}-r\sin \theta \frac{\partial u}{\partial x}
\end{align*}
I already have this, but I don't know how to continue.
 A: You identity for the Laplacian is not correct. Here we derive the correct identity  using some of your computations:

You already have computed
$$ \begin{align}
u_r &= u_x\cos\theta + u_y\sin\theta\\
u_{rr}&=u_{xx}\cos^2+u_{yy}\sin^2+2u_{xy}\cos\theta\,\sin\theta\\
u_\theta&=u_y r\cos\theta - u_x r\sin\theta
\end{align}$$
Now let's compute $u_{\theta\theta}$ using the product formula and the chain rule:
$$\begin{align}
u_{\theta\theta}&=\big(-r\sin\theta u_y +r\cos\theta(u_{yx}\partial_\theta x+ u_{yy}\partial_\theta y)\big)-\big(r\cos\theta u_x+r\sin\theta(u_{xx}\partial_\theta x+ u_{yx}\partial_\theta y)\big)\\
&=\big(-r u_y\sin\theta-r^2\cos\theta\sin\theta\, u_{xy}+r^2\cos^2\theta\,u_{yy}\big)-\big(r\ u_x\cos\theta-r^2\sin^2\theta\,u_{xx}+r^2\sin\theta\cos\theta\,u_{yx}\big)\\
&=-ru_y\sin\theta-ru_x\cos\theta +u_{xx}r^2\sin^2\theta +u_{yy}r^2\cos^2\sin\theta-u_{xy}r^2\cos\theta\,\sin\theta+u_{yx}r^2\cos\theta\,\sin\theta\\
&=-ru_y\sin\theta-ru_x\cos\theta +u_{xx}r^2\sin^2\theta +u_{yy}r^2\cos^2\sin\theta\\
&=-r u_r+u_{xx}r^2\sin^2\theta +u_{yy}r^2\cos^2\sin\theta
\end{align}$$
since $u_{xy}=u_{yx}$. Then
$$\frac{1}{r^2}u_{\theta\theta}=-\frac{u_r}{r}+u_{xx}\sin^2\theta +u_{yy}\cos^2\sin\theta
$$
Adding $u_{rr}$ and $u_{\theta\theta}$ gives
$$\begin{align}
u_{rr}+\frac{u_{\theta\theta}}{r^2}&=-\frac{u_r}{r}+u_{xx}(\cos^2\theta+\sin^2\theta)+u_{yy}(\cos^2+\sin^2\theta)\\
&=-\frac{u_r}{r}+u_{xx}+u_{yy}
\end{align}
$$
whence we obtain that
$$u_{xx}+u_{yy}=u_{rr}+\frac{u_{\theta\theta}}{r^2}+\frac{u_r}{r}$$
