= Solution
The proposed linear <change of variables> makes the two pairs of boundary lines constant-coordinate lines: $v=1,4$ and $u=0,7$. Thus the image of the <parallelogram> is the rectangle $0\le u\le7$, $1\le v\le4$.
The forward <Jacobian matrix> and its <determinant> are
$$
\frac{\partial(u,v)}{\partial(x,y)}=\begin{pmatrix}1&-3\\2&1\end{pmatrix},\qquad\det=7.
$$
Inverting gives $x=(u+3v)/7$, $y=(v-2u)/7$, so the area element is $dx\,dy=\tfrac17du\,dv$. The absolute inverse <Jacobian determinant>, rather than the forward determinant, belongs in the transformed area integral. Consequently
$$
\begin{aligned}
\iint_D\frac{x-3y}{2x+y}\,dx\,dy&=\frac17\int_0^7\int_1^4\frac uv\,dv\,du\\
&=\frac17\left[\frac{u^2}{2}\right]_0^7[\log v]_1^4,
\end{aligned}
$$
which yields $\boxed{\tfrac72\log4=7\log2}$. The transformed denominator is bounded below by one throughout the region, so no singular or improper integral is involved.
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