= Solution
For $x\ne0$ and $y\ne0$, put $v=y/x$. This is the standard substitution for a <homogeneous first-order differential equation>. Since $y'=v+xv'$, the equation becomes
$$
1+v^2-v(v+xv')=0,\qquad vv'=\frac1x.
$$
Taking <antiderivatives> yields $v^2=2\ln|x|+C$. The explicit general solution is therefore
$$
\boxed{y(x)=\pm x\sqrt{2\ln|x|+C},}
$$
on any <interval> where $x\ne0$ and $2\ln|x|+C>0$, with a fixed choice of sign. For a check that does not divide by $y$, set $w=y^2$ to get $w'-2w/x=2x$ and hence $w=x^2(2\ln|x|+C)$.
There is no missing zero solution: at $y=0$ with $x\ne0$, the original equation would require $x^2=0$. Also no real solution on an <interval> about $x=0$ can arise from this family, since $2\ln|x|+C$ becomes negative near $0$.
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