Solve the differential equation $y' = \frac{xy-y^2\ln{y}}{x^2\ln{x}- xy} $ The question is to solve the differential equation given below 
$$y' = \frac{xy-y^2\ln{y}}{x^2\ln{x}- xy} $$
First I tried simple  substitution but that didn't work , then I tried to cross multiply and arrange things in a manner so that I could (maybe ) solve the question , but that didn't help either.
 A: \begin{align*}
  \frac{dy}{dx} &= \frac{y(x-y\ln y)}{x(x\ln x-y)} \\
  &= \frac{x^2y^2 \left( \dfrac{1}{xy}-\dfrac{\ln y}{x^2} \right)}
          {x^2y^2 \left( \dfrac{\ln x}{y^2}-\dfrac{1}{xy} \right)} \\
  0 &=
  \left( \frac{1}{xy}-\frac{\ln y}{x^2} \right)dx+
  \left( \frac{1}{xy}-\frac{\ln x}{y^2} \right)dy \\
  \frac{\partial}{\partial y}
  \left( \frac{1}{xy}-\frac{\ln y}{x^2} \right)
  &= -\frac{1}{xy^2}-\frac{1}{x^2y} \tag{$\partial_y M$} \\
  \frac{\partial}{\partial x}
  \left( \frac{1}{xy}-\frac{\ln x}{y^2} \right)
  &= -\frac{1}{x^2y}-\frac{1}{xy^2} \tag{$\partial_x N$} \\
  0 &= \left( \frac{\ln y}{x}+\frac{\ln x}{y} \right)' \\
  k &= \frac{\ln y}{x}+\frac{\ln x}{y} \\
  kxy &= x\ln x+y\ln y \\
  e^{kxy} &= x^xy^y
\end{align*}
A: An addition to the implicit solution of Ng Chung Tak :
If we change $\enspace\displaystyle e^{kxy}=x^x y^y\enspace$ to $\enspace\displaystyle w=z^{1/z}\enspace$ with $\enspace\displaystyle z:=e^{kx}/y\enspace$ and $\enspace\displaystyle w:=x^{xe^{-kx}}\enspace$ then we know that there are solutions only possible for $\enspace\displaystyle 0<w\leq e^{1/e}$ . 
If we change $\enspace\displaystyle e^{kxy}=x^x y^y\enspace$ to $\enspace\displaystyle w=ze^z\enspace$ with $\enspace\displaystyle w:=-xe^{-kx}\ln x\enspace$ and $\enspace\displaystyle z:=-kx+\ln y\enspace$ then we can solve $\enspace\displaystyle w=ze^z\enspace$ for $\enspace z\enspace$ with Lamberts W-function .
