= Solution
{c}
Using the first <partial derivatives> from part (b),
$$
vg_u-ug_v=2uvf_x+(v^2-u^2)f_y=2yf_x-2xf_y.
$$
Also $uvg=yf$. Therefore the first <partial differential equation> becomes
$$
\boxed{2y\frac{\partial f}{\partial x}-2x\frac{\partial f}{\partial y}+yf=0.}
$$
For the second <partial differential equation>, the same <chain rule> gives
$$
ug_u-vg_v=(u^2+v^2)f_x,\qquad (u^4-v^4)g=2x(u^2+v^2)f.
$$
Away from the origin, cancellation yields $f_x+2xf=0$. Treating $y$ as fixed, an <integrating factor> gives $\partial_x(e^{x^2}f)=0$. Thus
$$
\boxed{f(x,y)=e^{-x^2}F(y),}
$$
where $F$ is an arbitrary sufficiently <differentiable function>. This is the local <general solution>, and the global form when $f$ is defined on the whole plane. If $f$ and its first <partial derivatives> are continuous at the origin, the reduced <partial differential equation> also holds there by <continuity>.
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