= Solution
With $v=x+y$, the <chain rule> gives
$$
\frac{dv}{dx}=1+\frac{dy}{dx}
=\frac{2(v-2)}{3v-4}.
$$
For $v\ne2$, this <separable differential equation> gives
$$
x=\int\frac{3v-4}{2(v-2)}\,dv
=\frac32v+\log|v-2|+C.
$$
Restoring $v$ and absorbing constants, the nonconstant branches satisfy
$$
\boxed{x+3y+2\log|x+y-2|=C}.
$$
The constant branch lost in division is $v=2$, giving the additional solution $\boxed{y=2-x}$. All branches must avoid $v=4/3$, where the original differential equation is undefined.
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