Given,
$$
\begin{aligned}
& \left(1+e^{x / y}\right) d x+e^{x / y}\left(1-\frac{x}{y}\right) d y=0 \\
& \Rightarrow \quad\left(1+e^{x / y}\right) d x=-e^{x / y}\left(1-\frac{x}{y}\right) d y \\
& \Rightarrow \quad \frac{d x}{d y}=\frac{-e^{x / y}\left(1-\frac{x}{y}\right)}{\left(1+e^{x / y}\right)}
\end{aligned}
$$
Putting $\quad x=v y$
Differentiating w.r.t. $y$,
$$
\begin{aligned}
\frac{d x}{d y} & =v \frac{d(y)}{d y}+y \frac{d v}{d y} \\
\Rightarrow \quad \frac{d x}{d y} & =v+y \frac{d v}{d y}
\end{aligned}
$$
Putting value of $\frac{d x}{d y}$ and $x=v y$ in Eq. (i)
$$
\begin{aligned}
\frac{d y}{d x} & =\frac{e^{-x / y}\left(1-\frac{x}{y}\right)}{1+e^{x / y}} \\
v+y \frac{d v}{d y} & =\frac{-e^v(1-v)}{1+e^v} \\
\Rightarrow \quad y \frac{d v}{d y} & =-\frac{e^v+v e^v}{1+e^v}-v \\
y \frac{d v}{d y} & =\frac{-e^v+v e^v-v-v e^v}{1-e^v} \\
\Rightarrow \quad\left[\frac{d v}{d y}\right. & =\frac{-\left[v+e^v\right]}{1+e^v} \\
\Rightarrow \quad\left[\frac{1+e^v}{v+e^v}\right] d v & =-\frac{d y}{y}
\end{aligned}
$$
Integrating both sides, we get
$$
\begin{aligned}
\int \frac{1+e^v}{v+e^v} d v & =-\int \frac{d y}{y} \\
\Rightarrow \quad \int \frac{1+e^v}{v+e^v} d v & =-\log y+\log c
\end{aligned}
$$
Put
$$
\begin{aligned}
v+e^v & =t \\
\left(1+e^v\right) d v & =d t
\end{aligned}
$$
Thus, $\quad \quad \quad \frac{d t}{t}=-\log y+\log c$
$$
\begin{aligned}
& \log t=-\log y+\log c \\
& \Rightarrow \quad \log \left(v+e^v\right)=-\log y+\log c \\
& \Rightarrow \quad \log y\left(v+e^v\right)=\log c \\
&
\end{aligned}
$$
Put value of $\quad v=\frac{x}{y}$
$$
\begin{aligned}
\log y\left(\frac{x}{y}+e^{x / y}\right) & =\log c \\
\Rightarrow \quad y\left(\frac{x}{y}+e^{x / y}\right) & =c \\
\Rightarrow \quad x+y e^{x / y} & =c
\end{aligned}
$$