Answer
$y'=\frac {y({y-e^{x/y})}}{{y^{2}-xe^{x/y}}}$
Work Step by Step
Differentiate $e^{x/y}=x-y$ with respect to $x$.
Apply chain rule.
$e^{x/y}\frac{d}{dx}(\frac{x}{y})=1-\frac{dy}{dx}$
Apply quotient rule.
$\frac{dy}{dx}=e^{x/y}\frac{d}{dx}(\frac{x}{y})+1$
$y'=e^{x/y}[\frac{1}{y}-\frac{x}{y^{2}}\frac{dy}{dx}]+1$
Hence, $y'=\frac {y({y-e^{x/y})}}{{y^{2}-xe^{x/y}}}$
