= Solution
The inverse <change of variables> is $y_1=uv$, $y_2=(1-u)v$, on $0<u<1$, $v>0$. Its absolute <Jacobian determinant> is
$$
\left|\det\begin{pmatrix}v&u\\-v&1-u\end{pmatrix}\right|=v.
$$
The <change of variables for probability densities> therefore gives
$$
\boxed{f_{U,V}(u,v)=\lambda^2v e^{-\lambda v}\mathbf1_{\{0<u<1,\ v>0\}}.}
$$
This factorizes on a product domain into a unit <uniform density> in $u$ and a shape-two <Gamma density> in $v$. Thus \b[$U$ and $V$ are independent], with $U\sim\operatorname{Uniform}(0,1)$ and $V\sim\operatorname{Gamma}(2,\lambda)$. This is the shape-$(1,1)$ case of <beta-gamma independence>.
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