Solution
= Solution
Use $U=Y$, $V=XY^2$, with <inverse element>[inverse] $y=u$, $x=v/u^2$. The Jacobian is $u^{-2}$, and the transformed support is $0<u<1$, $0<v<u^2$. Thus, for $0<v<1$,
$$
f_V(v)=\int_{\sqrt v}^1u^{-2}du=\frac1{\sqrt v}-1.
$$
Therefore $\boxed{f_V(v)=(v^{-1/2}-1)\mathbf1_{(0,1)}(v)}$. The integral over its support is $2-1=1$, checking normalization; there are no endpoint atoms.