Ok, I don't know how to go about this one at all.
Find dy/dx by implicit differentiation.
e^(x/y) = 8x - y
$\displaystyle \frac{d}{dx} \frac{x}{y} = \frac{1 \cdot y - xy^{\prime}}{y^2}$
So from Krizalid's post:
$\displaystyle \left ( \frac{y - xy^{\prime}}{y^2} \right ) \cdot e^{x/y}=8-y^{\prime}$
$\displaystyle y e^{x/y} - xy^{\prime}e^{x/y} = 8y^2 - y^2y^{\prime}$
$\displaystyle y^2y^{\prime} - xy^{\prime}e^{x/y} = 8y^2 - y e^{x/y}$
$\displaystyle (y^2 - xe^{x/y})y^{\prime} = 8y^2 - y e^{x/y}$
$\displaystyle y^{\prime} = \frac{8y^2 - y e^{x/y}}{y^2 - xe^{x/y}}$
(And you really need to use parenthesis!)
-Dan