Solution
= Solution
The zero <function> is a solution. On an interval where $y\ne0$, put
$$
z=\frac{y'}y.
$$
Then
$$
z'=\frac{yy''-(y')^2}{y^2}.
$$
The differential equation becomes
$$
xz'=z.
$$
Hence $z=Cx$, and integration gives
$$
\log|y|=\frac C2x^2+C_0.
$$
Thus all nonzero solutions are
$$
\boxed{y(x)=A\exp\left(\frac C2x^2\right)},
\qquad A\ne0,
$$
together with $\boxed{y\equiv0}$.