= Solution
<Independence of random variables> and the two unit-rate <exponential distributions> give
$$
f_{X,Y}(x,y)=e^{-x-y}\mathbf1_{\{x\geq0,\ y\geq0\}}.
$$
The <linear transformation>
$$
\begin{pmatrix}U\\V\end{pmatrix}
=\begin{pmatrix}6&8\\2&3\end{pmatrix}
\begin{pmatrix}X\\Y\end{pmatrix}
$$
has determinant $2$ and inverse
$$
x=\frac{3u-8v}{2},\qquad y=-u+3v.
$$
The inequalities $x,y\geq0$ become $8v/3\leq u\leq3v$. By the <change-of-variables formula for probability densities>,
$$
\boxed{f_{U,V}(u,v)=\frac12e^{-u/2+v}\mathbf1_{\{v\geq0,\ 8v/3\leq u\leq3v\}}}.
$$
The variables $U$ and $V$ are not independent: the support of their joint density is a wedge rather than a <Cartesian product> of one-dimensional supports. For example, conditioning on a positive value of $V$ constrains $U$ to a finite interval depending on that value.
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