= Solution
Put $y=ux$, so $y'=u+xu'$. On an interval with $x\ne0$, substitution and cancellation give
$$
x(u-1)\frac{du}{dx}=-(u^2+2u+2).
$$
Since $u^2+2u+2=(u+1)^2+1>0$, no constant-$u$ solution is lost when separating. Integrate
$$
\int\frac{u-1}{(u+1)^2+1}\,du=-\int\frac{dx}{x}
$$
to obtain
$$
\frac12\log((u+1)^2+1)-2\arctan(u+1)=-\log|x|+C.
$$
After restoring $u=y/x$, a local implicit <general solution> is
$$
\boxed{\frac12\log\left(x^2+(x+y)^2\right)
-2\arctan\left(\frac{x+y}{x}\right)=C}.
$$
The inverse tangent is chosen continuously on a solution branch. Using the continuous polar angle of $(x,x+y)$ extends the expression to charts through $x=0$ when the original equation permits it. A differentiable graph cannot pass through $y=x\ne0$, because there the original equation would require $5x=0$.
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