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