Answer
\[y(x)=\frac{Ce^x}{x}\]
Work Step by Step
\[\frac{dy}{dx}=\frac{y}{x}[\ln (xy)-1]\;\;\;\ldots (1)\]
Substitute $\: V=xy\;\;\;\ldots (2)$
Differentiate (2) with respect to $x$
\[\frac{dV}{dx}=y+x\frac{dy}{dx}\]
From (1)
\[\frac{dV}{dx}=y+x\left[\frac{y}{x}\{\ln (xy)-1\}\right]\]
\[\frac{dV}{dx}=y[1+\ln (xy)-1]\]
\[\frac{dV}{dx}=y\ln (xy)\]
From (2)
\[\frac{dV}{dx}=\frac{V}{x}\ln (V)\]
Separating variables
\[\frac{dV}{V\ln V}=\frac{1}{x}dx\]
Integrating, \[\int\frac{dV}{V\ln V}=\int\frac{1}{x}dx+\ln C_1\]
$\ln C_1$ is constant of integration
\[\ln |\ln V|=\ln |x|+\ln C_1\]
\[\ln |\ln V|=\ln |C_1 x|\]
\[\ln V=C_1 x\]
\[V=e^{C_1 x}\]
From (2)
\[xy=Ce^x\]
Where $C=e^{C_1}$
\[y=\frac{Ce^x}{x}\]
Hence \[y(x)=\frac{Ce^x}{x}.\]
