Answer
$y'=\frac{ye^{\frac{x}{y}}-y^2}{xe^{\frac{x}{y}}-y^2}$
Work Step by Step
$(e^{\frac{x}{y}}=x-y)'$
Note: To find $(e^{\frac{x}{y}})'$, we will need the quotient rule.
$(\frac{y-xy'}{y^2})e^{\frac{x}{y}}=1-y'$
$ye^{\frac{x}{y}}-xy'e^{\frac{x}{y}}=y^2-y'y^2$
Now, solve for $y'$.
$ye^{\frac{x}{y}}-y^2=xy'e^{\frac{x}{y}}-y'y^2$
$ye^{\frac{x}{y}}-y^2=y'(xe^{\frac{x}{y}}-y^2)$
$y'=\frac{ye^{\frac{x}{y}}-y^2}{xe^{\frac{x}{y}}-y^2}$
