Show that

$\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0$

$\Rightarrow\left(1+e^{\frac{x}{y}}\right) d x=-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y$

$\Rightarrow \frac{d x}{d y}=\frac{-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{1+e^{\frac{x}{y}}}$       ...(1)

Let $F(x, y)=\frac{-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{1+e^{\frac{x}{y}}}$.

$\therefore F(\lambda x, \lambda y)=\frac{-e^{\frac{\lambda x}{\lambda y}}\left(1-\frac{\lambda x}{\lambda y}\right)}{1+e^{\frac{\lambda x}{\lambda y}}}=\frac{-e^{\frac{x}{y}}\left(1-\frac{x}{y}\right)}{1+e^{\frac{x}{y}}}=\lambda^{0} \cdot F(x, y)$

Therefore, the given differential equation is a homogeneous equation.

To solve it, we make the substitution as:

x = vy

$\Rightarrow \frac{d}{d y}(x)=\frac{d}{d y}(v y)$

$\Rightarrow \frac{d x}{d y}=v+y \frac{d v}{d y}$

Substituting the values of $x$ and $\frac{d x}{d y}$ in equation (1), we get:

$v+y \frac{d v}{d y}=\frac{-e^{v}(1-v)}{1+e^{v}}$

$\Rightarrow y \frac{d v}{d y}=\frac{-e^{v}+v e^{v}}{1+e^{v}}-v$

$\Rightarrow y \frac{d v}{d y}=\frac{-e^{v}+v e^{v}-v-v e^{v}}{1+e^{v}}$

$\Rightarrow y \frac{d v}{d y}=-\left[\frac{v+e^{v}}{1+e^{v}}\right]$

$\Rightarrow\left[\frac{1+e^{v}}{v+e^{v}}\right] d v=-\frac{d y}{y}$

Integrating both sides, we get:

$\Rightarrow \log \left(v+e^{v}\right)=-\log y+\log \mathrm{C}=\log \left(\frac{\mathrm{C}}{y}\right)$

$\Rightarrow\left[\frac{x}{y}+e^{\frac{x}{y}}\right]=\frac{\mathrm{C}}{y}$

$\Rightarrow x+y e^{\frac{x}{y}}=\mathrm{C}$

This is the required solution of the given differential equation.