Answer
$y=x\ln\left(\frac{1}{C-\ln x}\right)$
Work Step by Step
Let $v=\frac{y}{x}$.
Then,
$y=xv$
$y'=v+v'x$
Substituting into the original differential equation,
$xy'=y+xe^{y/x}$
$x(v+v'x)=xv+xe^v$
$xv+v'x^2=xv+xe^v$
$v'x^2=xe^v$
$v'x=e^v$
Solve for $v$:
$\frac{dv}{dx}\cdot x=e^v$ (Separate the variables)
$e^{-v}dv=\frac{1}{x}dx$ (Integrate)
$\int e^{-v}dv=\int \frac{1}{x} dx$
$-e^{-v}=\ln x-C$
$e^v=\frac{1}{C-\ln x}$ (Take the logarithm)
$v=\ln\left(\frac{1}{C-\ln x}\right)$ (Back Substitute $v=\frac{y}{x}$)
$\frac{y}{x}=\ln\left(\frac{1}{C-\ln x}\right)$
$y=x\ln\left(\frac{1}{C-\ln x}\right)$