Solution
= Solution
The <inverse element>[inverse] change of variables is $x=u$, $y=uv$, whose <Jacobian determinant> is $u$. Since $X,Y$ are <independent>, the original <joint density> is $f(x)g(y)$. Thus
$$
\boxed{f_{U,V}(u,v)=|u|f(u)g(uv)\quad(u,v\in\mathbb R).}
$$
The transformation is nonsingular away from $u=0$; this line has probability zero because $X$ has a <probability density function>[density], so no extra mass is omitted.