## Calculus 8th Edition

$y'=\frac{y^{2}-ye^{x/y}}{{y^{2}-xe^{x/y}}}$
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, $y'=\frac{y^{2}-ye^{x/y}}{{y^{2}-xe^{x/y}}}$