$\frac{dy}{dx}=\frac{-e^y}{2x^2}+\frac{1}{x}$

$\frac{e^{-y}dy}{dx}=\frac{e^{-y}}{x}+\frac{-1}{2x^2}$

$\frac{-e^{-y}dy}{dx}+\frac{e^{-y}}{x}=\frac{1}{2x^2}$

Let $e^{-y}=t ...\left(\mathrm{i}\right)$

$e^{-y}\left(-1\right)\frac{dy}{dx}=\frac{dt}{dx}$

$\frac{dt}{dx}+\frac{t}{x}=\frac{1}{2x^2}$

$I.F{=}_e^{\int \frac{1}{x}dx}=e^{lnx}=x$

$tx=\int \frac{1}{2x^2}·xdx+C$

Using equation $\left(1\right)$
$e^{-y}x=\frac{1}{2}\ell nx+C$

Given, $y\left(e\right)=1$

$\Rightarrow e^{-1}e=\frac{1}{2}+C\Rightarrow C=\frac{1}{2}$

$e^{-y}x=\frac{1}{2}(1+\ell nx)$

Put $x=1$ then $y$ is $\Rightarrow y=\ell n2 \text{or} {\log }_{e}2$