$$y\frac{\partial u}{\partial x} - x\frac{\partial u}{\partial y} = c(x, y, u)$$ Express it in terms of $w(r,\theta) = u(r\cos\theta, r\cos\theta)$, $r$ and $\theta$.
I have $$\frac{\partial u}{\partial x} = \cos\theta\frac{\partial w}{\partial r} - \frac{\sin\theta}{r}\frac{\partial w}{\partial \theta}$$
$$\frac{\partial u}{\partial y} = \sin\theta\frac{\partial w}{\partial r} + \frac{\cos\theta}{r}\frac{\partial w}{\partial \theta}$$
So, the pde becomes
$$r\sin\theta\cos\theta\frac{\partial w}{\partial r} - \frac{r\sin^2\theta}{r}\frac{\partial w}{\partial \theta} - r\cos\theta\sin\theta\frac{\partial w}{\partial r} - \frac{r\cos^2\theta}{r}\frac{\partial w}{\partial \theta} = c(r\cos\theta, r\sin\theta, w(r,\theta))$$
Which reduces to:
$$-\frac{\partial w}{\partial \theta} = c(r\cos\theta, r\sin\theta, w(r,\theta))$$
Is this correct?
Yes. $\large\color{green}\checkmark$
Taking the previous answer in matrix form:
$$\begin{align}\begin{bmatrix}\dfrac{\partial w}{\partial r}\\\dfrac{\partial w}{\partial \theta}\end{bmatrix} &=\begin{bmatrix}\cos(\theta)&\sin(\theta)\\-r\sin(\theta)&r\cos(\theta)\end{bmatrix}\begin{bmatrix}\dfrac{\partial u}{\partial x}\\\dfrac{\partial u}{\partial y}\end{bmatrix}\end{align}$$
Applying matrix inversion produces: $$\begin{align}\begin{bmatrix}\dfrac{\partial u}{\partial x}\\\dfrac{\partial u}{\partial y}\end{bmatrix} &=\begin{bmatrix}\cos(\theta)&\sin(\theta)\\-r\sin(\theta)&r\cos(\theta)\end{bmatrix}^{-1}\begin{bmatrix}\dfrac{\partial w}{\partial r}\\\dfrac{\partial w}{\partial\theta}\end{bmatrix}\\ &=\begin{bmatrix}\cos(\theta)&-\sin(\theta)/r\\\sin(\theta)&\cos(\theta)/r\end{bmatrix}~\begin{bmatrix}\dfrac{\partial w}{\partial r}\\\dfrac{\partial w}{\partial\theta}\end{bmatrix}\end{align}$$
As you had. Substitution then leads to your final result. So it all checks out.