Switching to polar coordinates in differential equation Hi guys i need to help with this excercises. :)

If 
  $$\begin{cases}
x=r\cos \theta\\
y=r \sin\theta
\end{cases}$$
  prove that the equation:
  $$x^2\frac{\partial^2z}{\partial x^2}+2xy\frac{\partial^2z}{\partial x\partial y}+y^2\frac{\partial^2z}{\partial y^2}=0$$
  is equivalent to 
  $$r^2\frac{\partial^2z}{\partial \theta^2}=0.$$
OK I´VE TRIED

this could be helpful ? then, what do i make?
I tried of isolate  $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$
$$\frac{\partial z}{\partial x}=\frac{\partial z}{\partial r}cos\theta-\frac{\partial z}{\partial \theta}(\frac{1}{r})sin\theta$$and
$$\frac{\partial z}{\partial y}=\frac{\partial z}{\partial \theta}(\frac{1}{r})cos\theta+\frac{\partial z}{\partial r}sin\theta$$
mm and we know that
$$\frac{\partial z}{\partial r}=\frac{\partial z}{\partial x}cos\theta+\frac{\partial z}{\partial y}sin\theta$$
$$\frac{\partial z}{\partial \theta}=-\frac{\partial z}{\partial x}rsin\theta+\frac{\partial z}{\partial y}rcos\theta$$
but I do not get anything.
How can I do to get the result??
help.

You can convert the first differential equation from Cartesian coordinates to polar coordinates, or you can convert the second differential equation from polar coordinates to Cartesian coordinates.
I think the second way is slightly easier.
First, observe that by the product rule,
$$r \frac{\partial}{\partial r} \left(r \frac{\partial u}{\partial r}\right)
 = r \left(\frac{\partial r}{\partial r} \frac{\partial u}{\partial r}
           + r \frac{\partial}{\partial r}
                \left(\frac{\partial u}{\partial r}\right)\right)
 = r \frac{\partial u}{\partial r} + r^2 \frac{\partial^2 u}{\partial r^2},$$
that is,
$$ r^2 \frac{\partial^2 u}{\partial r^2}
 = r \frac{\partial}{\partial r} \left(r \frac{\partial u}{\partial r}\right)
   - r \frac{\partial u}{\partial r}. \tag 1$$
We can convert the right-hand side of this equation to Cartesian coordinates
as follows.
Let $x = r\cos\theta$ and $y = r\sin\theta$. Then for any $u$,
\begin{align}
r\frac{\partial u}{\partial r}
&= r\left(\frac{\partial u}{\partial x} \frac{\partial x}{\partial r}
         + \frac{\partial u}{\partial y} \frac{\partial y}{\partial r}\right) \\
&= r\left(\frac{\partial u}{\partial x} \cos\theta
         + \frac{\partial u}{\partial y} \sin\theta\right) \\
&= x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}.
\end{align}
Apply this formula,
$r\frac{\partial u}{\partial r}
 = x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}$,
with $r\frac{\partial u}{\partial r}$ in place of $u$:
\begin{align}
r\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}u\right)
&= x \frac{\partial}{\partial x} \left(r\frac{\partial}{\partial r}u\right)
   + y \frac{\partial}{\partial y} \left(r\frac{\partial}{\partial r}u\right) \\
&= x \frac{\partial}{\partial x} 
 \left(x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}\right)
   + y \frac{\partial}{\partial y}
 \left(x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y}\right)
\end{align}
Working out the various terms,
\begin{align}
x \frac{\partial}{\partial x} \left(x \frac{\partial u}{\partial x}\right)
 &= x \frac{\partial u}{\partial x} + x^2 \frac{\partial^2 u}{\partial x^2}, \\
x \frac{\partial}{\partial x} \left(y \frac{\partial u}{\partial y}\right)
 &= xy \frac{\partial^2 u}{\partial x \partial y}, \\
y \frac{\partial}{\partial y} \left(x \frac{\partial u}{\partial x}\right)
 &= xy \frac{\partial^2 u}{\partial x \partial y}, \text{ and}\\
y \frac{\partial}{\partial y} \left(y \frac{\partial u}{\partial y}\right)
 &= y \frac{\partial u}{\partial y} + y^2 \frac{\partial^2 u}{\partial y^2}.
\end{align}
Therefore
\begin{align}
r\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}u\right)
&= x^2 \frac{\partial^2 u}{\partial x^2}
  + 2 xy \frac{\partial^2 u}{\partial x \partial y}
  + y^2 \frac{\partial^2 u}{\partial y^2}
  + x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} \\
&= x^2 \frac{\partial^2 u}{\partial x^2}
  + 2 xy \frac{\partial^2 u}{\partial x \partial y}
  + y^2 \frac{\partial^2 u}{\partial y^2}
  + r \frac{\partial}{\partial r} u,
\end{align}
that is,
$$ r\frac{\partial}{\partial r}\left(r\frac{\partial}{\partial r}u\right)
    -  r \frac{\partial u}{\partial r}
  = x^2 \frac{\partial^2 u}{\partial x^2}
    + 2 xy \frac{\partial^2 u}{\partial x \partial y}
    + y^2 \frac{\partial^2 u}{\partial y^2}. \tag 2$$
Taking Equations $(1)$ and $(2)$ together, we find that
$$ r^2 \frac{\partial^2 u}{\partial r^2}
  = x^2 \frac{\partial^2 u}{\partial x^2}
    + 2 xy \frac{\partial^2 u}{\partial x \partial y}
    + y^2 \frac{\partial^2 u}{\partial y^2}.$$
This is a little stronger than the fact that was to be proved, but it
certainly does imply that the equations
$x^2 \frac{\partial^2 u}{\partial x^2}
    + 2 xy \frac{\partial^2 u}{\partial x \partial y}
    + y^2 \frac{\partial^2 u}{\partial y^2} = 0$
and $r^2 \frac{\partial^2 u}{\partial r^2} = 0$ are equivalent.
