Solution
= Solution
Work where $x\ne0$ and $f$ is differentiable. Set $q=y/x$. The <chain rule> gives
$$
w_x=f'(q)\left(-\frac y{x^2}\right),\qquad w_y=f'(q)\frac1x.
$$
Consequently
$$
\boxed{xw_x+yw_y=-\frac yx f'(q)+\frac yx f'(q)=0.}
$$
Geometrically, a radial rescaling of both coordinates leaves $y/x$ unchanged, so the function has zero rate of change along radial dilations. No value or derivative at $x=0$ is implied by this representation.