The solution  $x\frac{dy}{dx}=y+x{e}^{\frac{y}{x}}$ with y(1) =0

Put y=vx then $\frac{dy}{dx}=v+x\frac{dv}{dx}$

Given eq’n  $x\frac{dy}{dx}=y+x{e}^{\frac{y}{x}}$

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

$v+x\frac{dv}{dx}=v+e^v$

$\int e^{-v}dv=\int \frac{dx}{x}\Rightarrow e^{-v}=\log x-c$

$\begin{array}{l}\log x+e^{-v}=c\\ y\left(1\right)=0\end{array}$

$\log \left(1\right)+e^0=c\Rightarrow c=1$

The required equation is $e^{-\frac{y}{x}}+\log x=1$