Help in solving a differential equation $$\frac{dy}{dx} = \frac{-(xy\cdot e^x \ln(y)+ye^y)}{xy\cdot e^y\ln(x) + xe^x}$$ This equation cannot be homogenized. I tried to separate the variables and find suitable differential coefficients but I could not do so. I can see some symmetry in the top and bottom functions. Factorizing out x & y from the equation, I obtain $$\frac{dy}{dx} = \frac{-y(x\cdot e^x \ln(y)+e^y)}{x(y\cdot e^y\ln(x) + e^x)}$$ This gives us $f(x,y)$ in the numerator and $f(y,x)$ in the denominator. I am unsure if this helps to solve the equation and am unable to proceed from here. Any help is appreciated!

Substitute: $$ \begin{cases}w=\ln y \implies y=e^w \\\ v=\ln x \implies x=e^v \end{cases} $$ $$\frac{dy}{dx} = \frac{-y(x\cdot e^x \ln(y)+e^y)}{x(y\cdot e^y\ln(x) + e^x)}$$ $$\frac{dw}{dv} = \frac{-(e^v\cdot e^{e^v} w+e^{e^w})}{(e^w\cdot e^{e^w}v + e^{e^v})}$$ $${dw}({(e^w\cdot e^{e^w}v + e^{e^v})}) = ({-(e^v\cdot e^{e^v} w+e^{e^w})})dv$$ $$e^we^{e^w}v{dw} + e^{e^v}{dw} = -e^ve^{e^v}w dv-e^{e^w}dv$$ This differential I think is exact $$e^we^{e^w}v{dw} + e^{e^v}{dw} = -e^ve^{e^v}w dv-e^{e^w}dv$$ $$v{de^{e^w}} + e^{e^v}{dw} +w de^{e^v}+e^{e^w}dv=0$$ $${dve^{e^w}} + dwe^{e^v}=0$$ Integration: $${ve^{e^w}} + we^{e^v}=K$$ Finally substitute back $x$ and $y$ $${e^y\ln x} + e^{x}\ln y=K$$ Take the exponetial on both sides if you want Maple solution posted by Dr. Sonnhard Graubner $$x^{e^y}y^{e^x}=C$$