Solve Linear PDE via a change of variables

$\displaystyle

\omega(r,\theta)=u(r cos (\theta),r sin (\theta))

$ for some function u(x,y)

$\displaystyle

\displaystyle y \frac{\partial u}{\partial x}-x\frac{\partial u}{\partial y}=1

$

Rewrite the PDE and IC's in terms of $\displaystyle \omega,r, \theta$ by expressing

$\displaystyle

\displaystyle \frac{\partial \omega}{\partial r}$ and $\displaystyle

\displaystyle \frac{\partial \omega}{\partial \theta}$in terms of $\displaystyle

\displaystyle \frac{\partial u}{\partial x}$ and $\displaystyle \displaystyle \frac{\partial u}{\partial y}

$ and solve.

My attempt: I believe the first equation is of the form

$\displaystyle

\displaystyle \omega(r,\theta) = u(x(r,\theta),y(r,\theta))$ therefore$\displaystyle \begin{cases}

\frac{\partial \omega}{\partial r} = \frac{\partial \omega}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial \omega}{\partial y} \frac{\partial y}{\partial r}\\\frac{\partial \omega}{\partial \theta} = \frac{\partial \omega}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial \omega}{\partial y} \frac{\partial y}{\partial \heta}\end{cases} $ but they are not in terms of $\displaystyle

\displaystyle \frac{\partial u}{\partial x}$ and $\displaystyle \displaystyle \frac{\partial u}{\partial y}

$

Have I gone one step too far?