Given differential equation is
$$
\begin{gathered}
x \frac{d y}{d x}=y+x e^{y / x} \\
\frac{d y}{d x}=\frac{y}{x}+e^{y / x}
\end{gathered}
$$
It is a homogeneous differential equation.
$$
\begin{array}{llrl}
& \therefore \text { Put } y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \\
& \therefore & v+x \frac{d v}{d x}=\frac{v x}{x}+e^{v x / x} \\
\Rightarrow & v+x \frac{d v}{d x} & =v+e^v \\
\Rightarrow & x \frac{d v}{d x} & =e^v
\end{array}
$$
$$
\Rightarrow \quad e^{-v} d v=\frac{1}{x} d x
$$
On integrating both sides, we get
$$
\begin{aligned}
-e^{-v} & =\log x+c \\
-e^{-y / x} & =\log x+c
\end{aligned}
$$
Given, $y(1)=0$
$$
\begin{array}{rlrl}
& \therefore & e^{-0 / 1} & =\log 1+c \\
& & -1 & =0+c \Rightarrow c=-1 \\
\therefore & -e^{-y / x} & =\log x-1] \\
\Rightarrow & & 1 & =\log x+e^{-y / x}
\end{array}
$$