Solve the following differential equation: (1+e^y / x) d y+e^y / x(1-y/x) d x=0

Given differential equation can be written as \[\frac{d y}{d x}=\frac{e^{y / x}\left(\dfrac{y}{x}-1\right)}{1+e^{y / x}} \] put \(\displaystyle y=v x\) and \(\displaystyle \frac{d y}{d x}=v+x \frac{d v}{d x}\), we get \[v+x \frac{d v}{d x}=\frac{e^{v}(v-1)}{1+e^{v}} \] \(\displaystyle x \frac{d v}{d x}=\frac{-e^{v}-v}{1+e^{v}}\) \[\int \frac{\left(e^{v}+1\right)}{e^{v}+v} d v=-\int \frac{d x}{x} \] \(\displaystyle \log \left|e^{v}+v\right|=-\log |x|+\log C\) \[e^{v}+v=\frac{C}{x} \] \(\displaystyle e^{y / x}+\frac{y}{x}=\frac{C}{x} \quad\) or \(\displaystyle x e^{y / x}+y=C\)