# How do you implicitly differentiate 9=e^y/e^x+y^3-xy?

$\frac{d}{\mathrm{dx}} \left({e}^{y} / {e}^{x} + {y}^{3} - x y = 9\right)$
$0 = - {e}^{y} / {e}^{x} + {e}^{y} / {e}^{x} \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - x \frac{\mathrm{dy}}{\mathrm{dx}} - y$
dy/dx=(e^y/e^x+y)/(e^y/e^x+3y^2-x