The solution of  $(1+{\mathrm{e}}^{\mathrm{x}/\mathrm{y}})\mathrm{dx}+{\mathrm{e}}^{\mathrm{x}/\mathrm{y}}\left(1-\frac{\mathrm{x}}{\mathrm{y}}\right)\mathrm{dy}=0$ is

$\begin{array}{l}Given(1+e^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.})dx+e^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}(1-\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.)dy=0\\ \Rightarrow (1+e^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.})dx=(\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.-1)e^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}dy\\ \Rightarrow \frac{dx}{dy}=\frac{(\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.-1)e^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}}{1+e^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}}\\ Putx=vy\Rightarrow \frac{dx}{dy}=v+y.\frac{dv}{dy}\\ \Rightarrow v+y.\frac{dv}{dy}=\frac{(v-1)e^v}{1+e^v}\\ \Rightarrow y\frac{dv}{dy}=\frac{v{e}^v-e^v}{1+e^v}-v\\ =\frac{v{e}^v-e^v-v-e^{v}v}{1+e^v}\end{array}$

$\begin{array}{l}\Rightarrow \int \frac{e^v+1}{e^v+v}=-\int \frac{1}{y}dy\\ \Rightarrow \log (e^v+v)=-\log y+\log c\\ \Rightarrow e^v+v=\frac{c}{y}\\ \Rightarrow y(e^v+v)=c\\ \Rightarrow y(e^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}+\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.)=c\\ \Rightarrow y{e}^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$y$}\right.}+x=c\end{array}$