Showing that two equations are equal using chain rule. 
Let $u = f(x,y)$, with $x= r \cos\theta$, $y =r\sin\theta$.
  Show that $$\left(\frac{\partial u}{\partial r}\right)^2+\frac{1}{r^2}\left(\frac{\partial u}{\partial \theta}\right)^2=\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2$$

How do I solve this question when the equation of $u$ is not given? I know that to get $\left(\dfrac{du}{dr}\right)^{2}$, I would first need to get $\left(\dfrac{du}{dx}\right)$ and $\left(\dfrac{du}{dy}\right)$ because
$$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\cdot\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\cdot\frac{\partial y}{\partial r}$$ 
Also what do the squares on the partial derivatives imply? Do they to just do a simple square of the values or do they mean more? Thanks in advance!
 A: $x= r \cos\theta$ , $y= r \sin\theta$
Using

$$\frac{\partial u}{\partial \theta}=\frac{\partial u}{\partial x}\cdot\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial y}\cdot\frac{\partial y}{\partial \theta}\quad\text{and} \quad\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\cdot\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\cdot\frac{\partial y}{\partial r}$$

We have
$$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\cdot\cos\theta+\frac{\partial u}{\partial y}\cdot\sin\theta$$
$$\frac{\partial u}{\partial \theta}=\frac{\partial u}{\partial x}\cdot(-r\sin\theta)+\frac{\partial u}{\partial y}\cdot(r\cos\theta)$$
By squaring these two equations we get
$$\begin{align}\left(\frac{\partial u}{\partial r}\right)^2=&\left(\frac{\partial u}{\partial x}\cdot\cos\theta+\frac{\partial u}{\partial y}\cdot\sin\theta\right)^2\\
=&\left(\frac{\partial u}{\partial x}\right)^2\cdot\cos^2\theta+\left(\frac{\partial u}{\partial y}\right)^2\cdot\sin^2\theta+2\left(\frac{\partial u}{\partial x}\right)\left(\frac{\partial u}{\partial y}\right)\sin\theta\cos\theta\tag{1}\\
\end{align}$$
$$\begin{align}\frac{1}{r^2}\left(\frac{\partial u}{\partial \theta}\right)^2&=\left(\frac{\partial u}{\partial x}\cdot(-\cos\theta)+\frac{\partial u}{\partial y}\cdot\sin\theta\right)^2\\
&=\left(\frac{\partial u}{\partial x}\right)^2\cdot\sin^2\theta+\left(\frac{\partial u}{\partial y}\right)^2\cdot\cos^2\theta-2\left(\frac{\partial u}{\partial x}\right)\left(\frac{\partial u}{\partial y}\right)\sin\theta\cos\theta\tag{2}\\
\end{align}$$
Adding $(1)$ and $(2)$ we get 
$$\left(\frac{\partial u}{\partial r}\right)^2+\frac{1}{r^2}\left(\frac{\partial u}{\partial \theta}\right)^2=\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2$$
