The partial differential equation is given by $y\frac{\partial u}{\partial x} - x\frac{\partial u}{\partial y} = c(x, y, u)$
Define $w(r,\theta) = u(r\cos\theta, r \sin\theta)$ for $r>0$, $\theta \in R$
What steps do I take to begin expressing: $\frac{\partial w}{\partial r}$ and $\frac{\partial w}{\partial \theta}$ in terms of $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$?
Does the answer work out to?
$\frac{\partial w}{\partial r} = \cos\theta\frac{\partial u}{\partial x} + \sin\theta\frac{\partial u}{\partial y}$
$\frac{\partial w}{\partial \theta} = -r\sin\theta\frac{\partial u}{\partial x} + r\cos\theta\frac{\partial u}{\partial y}$
Indeed, since you are taking: $x(r,\theta)=r\cdot\cos(\theta)$ and $y(r,\theta)=r\cdot\sin(\theta)$, then all you need is to apply the chain rule and evaluate the derivatives:$$\begin{align}\dfrac{\partial w}{\partial r}&=\dfrac{\partial x}{\partial r}\dfrac{\partial u}{\partial x}+\dfrac{\partial y}{\partial r}\dfrac{\partial u}{\partial y}\\&=\cos(\theta)\cdot\dfrac{\partial u}{\partial x}+\sin(\theta)\cdot\dfrac{\partial u}{\partial y}\\[1ex]\dfrac{\partial w}{\partial \theta}&=\dfrac{\partial x}{\partial \theta}\dfrac{\partial u}{\partial x}+\dfrac{\partial y}{\partial \theta}\dfrac{\partial u}{\partial y}\\&=-r\sin(\theta)\cdot\dfrac{\partial u}{\partial x}+r\cos(\theta)\cdot\dfrac{\partial u}{\partial y}\end{align}$$
PS: of course, $\cos(\theta)=x/r,~ \sin(\theta)=y/r$ and $r^2=x^2+y^2$, so...