= Solution
The proposed <change of variables> gives $u=x-3y$, $v=2x+y$, with inverse $x=(u+3v)/7$, $y=(v-2u)/7$. The <Jacobian determinant> is
$$
\frac{\partial(u,v)}{\partial(x,y)}=7,\qquad \left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\frac17.
$$
The two lines of slope $-2$ become $v=1,4$, while the two lines of slope $1/3$ become $u=0,7$. Thus the parallelogram becomes $0\leq u\leq7$, $1\leq v\leq4$. Since $v$ stays positive, the transformed integrand has no singularity there. The <double integral> is
$$
\boxed{\iint_D\frac{x-3y}{2x+y}\,dx\,dy=\frac17\int_0^7u\,du\int_1^4\frac{dv}{v}=\frac72\ln4=7\ln2.}
$$
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