Equation is $\displaystyle xy' = y \ln \frac{y}{x}$.

I have tried to substitute $\displaystyle v = \frac{y}{x}$ and I am stuck at $\displaystyle \frac{1}{dx}x = v (\ln v - 1)\frac{1}{dv}$

To be exact I do not know where to go from there in calculus sense.