= Solution
The <change of variables> has inverse $x=(u+3v)/7$, $y=(v-2u)/7$, with absolute inverse <Jacobian determinant> $1/7$. The two lines $y=-2x+1, -2x+4$ become $v=1,4$, and $y=x/3,(x-7)/3$ become $u=0,7$. Hence the parallelogram maps to a rectangle and
$$
\boxed{\iint_D\frac{x-3y}{2x+y}\,dx\,dy=\frac17\int_0^7\int_1^4\frac uv\,dv\,du=\frac72\ln4=7\ln2.}
$$
The denominator $v$ is bounded away from zero throughout the rectangle, so this transformation introduces no improper-integral issue.
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