The differential equation $\frac{d y}{d x}+\frac{y^2}{x^2}=\frac{y}{x}$ has the solution

$x=y(\log x+C)$

We have, $\frac{d y}{d x}+\frac{y^2}{x^2}=\frac{y}{x}$

Putting $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$, we get

$v+x \frac{d v}{d x}+v^2=v \Rightarrow x \frac{d v}{d x}=-v^2 \Rightarrow-\frac{1}{v^2} d v=\frac{1}{x} d x$

On integrating, we get

$\frac{1}{v}=\log x+C \Rightarrow x=y(\log x+C)$