= Solution
The real logarithm requires $x>0$. The identically zero function is a solution and must be retained before dividing by $y$. For a nonzero solution, divide by $x$ and make the <Bernoulli differential equation> substitution $u=1/y$. This gives
$$
u'-\frac ux=-x\ln x,\qquad \left(\frac ux\right)'=-\ln x.
$$
Integration yields $u/x=-x\ln x+x+C$. Consequently
$$
\boxed{y\equiv0\quad\text{or}\quad y(x)=\frac1{Cx+x^2(1-\ln x)},\qquad x>0,}
$$
with each nonzero branch restricted to an interval where its denominator is nonzero. The right-hand side of the first-order equation is locally Lipschitz in $y$ for $x>0$, so a nonzero solution cannot cross the zero solution at a finite positive $x$.
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