\(x\dfrac{\mathrm dy}{\mathrm dx}=y\ln(xy)\)
Let \(v=xy\), so that \(\dfrac{\mathrm dv}{\mathrm dx}=y+x\dfrac{\mathrm dy}{\mathrm dx}\), or \(\dfrac{\mathrm dv}{\mathrm dx}-\dfrac vx=x\dfrac{\mathrm dy}{\mathrm dx}\). Then you have
\(\dfrac{\mathrm dv}{\mathrm dx}-\dfrac vx=\dfrac vx\ln v\)
\(\dfrac{\mathrm dv}{v(\ln v+1)}=\dfrac{\mathrm dx}x\)
Integrate both sides to get
\(\ln|\ln v+1|=\ln|x|+C\)
\(\ln v+1=Cx\)
\(v=e^{Cx-1}\)
then back-substitute to find the solution for \(y\).
\(xy=e^{Cx-1}\implies y=\dfrac{e^{Cx-1}}x\)
