= Solution
{c}
The <joint density> of $X,Y$ is $\lambda^2e^{-\lambda(x+y)}$ for positive $x,y$. Inverting the transformation gives $x=uv$, $y=u(1-v)$, with absolute <Jacobian determinant> $u$ and domain $u>0$, $0<v<1$. The <change of variables for probability densities> gives
$$
\boxed{f_{U,V}(u,v)=\lambda^2u e^{-\lambda u}\mathbf1_{\{u>0,\ 0<v<1\}},\qquad
f_U(u)=\lambda^2u e^{-\lambda u}\mathbf1_{\{u>0\}}.}
$$
The <joint density> factors into the gamma density of $U$ and the uniform density of $V$ on $(0,1)$, so these two transformed variables are independent.
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