By substituting $y=v x$, the solution of the differential equation
$\frac{d y}{d x}=\frac{x^2+y^2}{x y}$, is 

$\frac{y^2}{2 x^2}=\log x+C$

Substituting $y=v x$ and $\frac{d y}{d x}=v+x \frac{d v}{d x}$ in the given differential equation, we get

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

On integrating, we get

$\frac{v^2}{2}=\log x+C \Rightarrow \frac{y^2}{2 x^2}=\log x+C$