= Solution
Work on an <interval> with $x\ne0$. The <homogeneous first-order differential equation> becomes a <separable differential equation> under $v=y/x$: substituting $y=xv$ gives $v+xv'=v^2+v$, hence $xv'=v^2$.
For $v\ne0$, <integration> gives $-1/v=\ln|x|+C$. Renaming the arbitrary constant produces
$$
\boxed{y(x)=\frac{x}{C-\ln|x|},\qquad\text{or}\qquad y(x)\equiv0.}
$$
For the nonzero family the denominator must remain nonzero, so take any connected <interval> avoiding both $x=0$ and $|x|=e^C$. The zero solution must be restored after division by $v^2$; by the <Picard-Lindelöf theorem> away from $x=0$, a solution cannot switch between zero and a nonzero branch there.
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