= Solution
The inverse <linear transformation> is
$$
x=2u-v,\qquad y=v-u,
\qquad
\left|\det\frac{\partial(x,y)}{\partial(u,v)}\right|
=\left|\det\begin{pmatrix}2&-1\\-1&1\end{pmatrix}\right|=1.
$$
The 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> yields
$$
\boxed{f_{U,V}(u,v)=e^{-u}\mathbf1_{\{u>0,\ u<v<2u\}}.}
$$
Although 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 $V$.
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