# How do you implicitly differentiate  ysinx^2 = xe^-ysin(e^y)?

$\frac{\mathrm{dy}}{\mathrm{dx}} . \sin {x}^{2} + 2 x y \cos {x}^{2} = {e}^{-} y \sin {e}^{y} + x \left(- {e}^{-} y . \frac{\mathrm{dy}}{\mathrm{dx}} . \sin {e}^{y} + \frac{\mathrm{dy}}{\mathrm{dx}} \cos {e}^{y}\right)$