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.
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 $$u_{xx}+u_{yy}=u_{rr}+\frac{u_{\theta\theta}}{r^2}+\frac{u_r}{r}$$