Rewrite the ODE as
\(y\log y\,\mathrm dx+(x-\log y)\,\mathrm dy=0\iff y\log y\dfrac{\mathrm dx}{\mathrm dy}+x=\log y\)
\(\dfrac{\mathrm dx}{\mathrm dy}=\dfrac1{y\log y}x=\dfrac1y\)
so that it is now linear in \(x\). An integrating factor would
\(\mu(y)=\exp\left(\displaystyle\int\frac{\mathrm dy}{y\log y}\right)=e^{\log(\log y)}=\log y\)
Multiply both sides by \(\mu(y)\) to get
\(\log y\dfrac{\mathrm dx}{\mathrm dy}=\dfrac1yx=\dfrac{\log y}y\)
\(\dfrac{\mathrm d}{\mathrm dy}[x\log y]=\dfrac{\log y}y\)
\(x\log y=\displaystyle\int\frac{\log y}y\,\mathrm dy\)
\(x\log y=\dfrac12\log^2y+C\)
\(x=\dfrac12\log y+\dfrac C{\log y}\)
