$\left(1+e^{\frac{x}{y}}\right) d x+\left(1-\frac{x}{y}\right) e^{x / y} d y=0$
$\Rightarrow \quad\left(1+e^{x / y}\right) d x=-e^{x / y}\left(1-\frac{x}{y}\right) d y$
$\Rightarrow \quad \frac{d x}{d y}=\frac{-e^{x / y}(1-x / y)}{\left(1+e^{x / y}\right)}$ ...(i)
This is homogeneous differential equation.
So, put $\quad x=v y$
$\frac{d x}{d y}=v+y \frac{d v}{d y}$ ....(ii)
Therefore, $v+y \frac{d v}{d y}=\frac{-e^{v} (1-v)}{1+e^{v}}$
[from Eqs.(i) and (ii)]
$\Rightarrow \quad y \frac{d v}{d y}=\frac{-e^{v}+ve^{v}}{1+e^{v}}-v$
$=\frac{-e^{v}+ve^{v}-v\left(1+e^{v}\right)}{1+e^{v}}$
$=-e^{v}+v e^{v}-v-v e^{v}$
$\Rightarrow \quad y \frac{d v}{d y}=\frac{-\left(e^{v}+v\right)}{1+e^{v}}$
$\Rightarrow \quad \frac{1+e^{v}}{v+e^{r}} d v=-\frac{d y}{y} \quad$ [variable separation $]$
$\Rightarrow \quad \int \frac{1+e^{v}}{v+e^{v}} d v=-\int \frac{d y}{y} \quad$ [by integration $] \ldots$ (iii)
Let $\left.v+e^{v}=t \Rightarrow (1+e^{v}\right) d v=d t$
$\int \frac{1+e^{v}}{v+e^{v}} d v=\int \frac{d t}{t}=\log t=\log \left(v+e^{v}\right)$ ...(iv)
$\Rightarrow \log \left(v+e^{v}\right)=-\log y+\log C$
[from Eqs. (iii) and (iv)] $\Rightarrow \log \left(v+e^{v}\right)+\log y=\log C$
$\Rightarrow \quad y\left(v+e^{v}\right)=C$
$y\left(\frac{x}{y}+e^{x / y}\right)=C$
$y\left(\frac{x+ye^{x / y}}{y}\right)=C$
$\Rightarrow \quad x+y e^{x / y}=C$