Answer
$y=ke^{\frac{\ln^2{x}}{2}}$
Work Step by Step
$y\ln{x}-xy'=0$
$y\ln{x}=xy'$
$\frac{\ln{x}}{x}=\frac{y'}{y}$
$\int\frac{dy}{y}=\int\frac{\ln{x}}{x}dx$
$\ln{y}=\frac{\ln^2{x}}{2}+C$
$e^{\ln{y}}=y=e^{\frac{\ln^2{x}}{2}+C}=e^{\frac{\ln^2{x}}{2}}e^C$
Let $k=e^C$ then
$y=ke^{\frac{\ln^2{x}}{2}}$
