 Solution

The inverse linear transformation isThe positive support $x>0,y>0$ is exactly $u>0$ and $u<v<2u$. The independence and rate-one exponential distributions give $f_{X,Y}(x,y)=e^{-x-y}$ there. Since $x+y=u$, the change of variables for probability densities yieldsAlthough the density's value depends only on $u$, its support does not factor as a product of fixed intervals. Thus this expression does not imply independence of $U$ and $V$.