Solving the differential equation: $y'x\log y =1$ we easly find : $$ y(\log y-1)=\log x +c $$
I search an explicit solution $y=f(x)$ and WolframAlpha gives: $$ y=\dfrac{\log x+c}{W\left( \dfrac{\log x +c}{e}\right)} $$
Where $W$ is the Lambert function. I know that this function is defined such that $W(ze^z)=z$, but I don't see how this can give the Wolfram result.