If $x \frac{d y}{d x}=y(\log y-\log x+1)$, then the solution of the equation is

$\log \left(\frac{y}{x}\right)=C x$

We have,

$x \frac{d y}{d x}=y \log \left(\frac{y}{x}\right)+y$

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

$v+x \frac{d v}{d x}=v \log v+v$

$\Rightarrow \frac{1}{v \log v} d v=\frac{1}{x} d x$

$\Rightarrow \log (\log v)=\log x+\log C$           [On integrating]

$\Rightarrow \log v=C x$

$\Rightarrow \log \left(\frac{y}{x}\right)=C x$, which is the required solution.