 Solution

The logarithm requires $x>0$ for real-valued solutions. Put $v=xy$. Since $v^{\prime }=x{y}^{\prime }+y$, the equation becomes $v^{\prime }=v^2\log x$. For a nonzero branch, separation givesRenaming the constant and restoring $y=v/x$ givesDividing by $v^2$ would omit the equilibrium $v=0$, so the full solution set also includes $\boxed{y\equiv 0}$. The nonzero branches are defined on connected intervals with $x>0$ and $C+x-x\log x\ne 0$. 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.