Answer
$y=xe^{Cx+1}$
Work Step by Step
Given:
$$xy'+ y\ln x=y \ln y$$
Let $y=vx \rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Subtituting:
$$x(v+xv')+ vx\ln x=vx \ln vx$$
$$x(v+xv')+ vx\ln x=vx \ln v+ vx \ln x$$
$$x(v+xv')=vx \ln v$$
$$xv+x^2v'=vx \ln v$$
$$x^2v'=vx\ln v - xv$$
$$xv'=v\ln v - v$$
$$xv'=v(\ln v -1)$$
$$\frac{dv}{v(\ln v -1)}=\frac{1}{x}dx$$
Integrating left side:
$\int \frac{dv}{v(\ln v -1)}$
Let $v=u \rightarrow \frac{1}{v}dv=du$
then $\int \frac{du}{u-1}=\ln (u-1)=\ln(\ln v-1)$
and $\ln(\ln v-1)=\ln x + C$
Subtitute when $y=vx \rightarrow y=\frac{y}{x} \rightarrow \ln\frac{y}{x}-1=e^{\ln x + C}+C$
$y=xe^{Cx+1}$
