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

The given equation is:
$$\frac{dy}{dx}=\frac{y}{x}(\log \frac{y}{x}+1)$$
This is a homogeneous differential equation. Let $y=vx⟹\frac{dy}{dx}=v+x\frac{dv}{dx}$.
$$v+x\frac{dv}{dx}=v(\log v+1)$$
$$v+x\frac{dv}{dx}=v\log v+v$$
$$x\frac{dv}{dx}=v\log v$$
Separating variables:
$$\frac{dv}{v\log v}=\frac{dx}{x}$$
Integrating both sides:
$$\int \frac{dv}{v\log v}=\int \frac{dx}{x}$$
Let $\log v=t⟹\frac{1}{v}dv=dt$.
$$\ln (\ln v)=\ln x+\ln c=\ln (cx)$$
$\ln v=cx⟹\log (y/x)=cx$.