= Solution
The logarithm requires $x>0$ for real-valued solutions. Put $v=xy$. Since $v'=xy'+y$, the equation becomes $v'=v^2\log x$. For a nonzero branch, separation gives
$$
-\frac1v=\int\log x\,dx=x\log x-x+C_0.
$$
Renaming the constant and restoring $y=v/x$ gives
$$
\boxed{y=\frac1{x(C+x-x\log x)}}.
$$
Dividing by $v^2$ would omit the equilibrium $v=0$, so the full solution set also includes $\boxed{y\equiv0}$. The nonzero branches are defined on connected intervals with $x>0$ and $C+x-x\log x\ne0$. A regular solution cannot switch from a nonzero branch to zero at an ordinary point, because the right-hand side is locally Lipschitz in $v$ there. These domain restrictions are part of the full real solution.
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