Implicit differentiation of an exponential equation

Differentiate both sides with respect to $x$. For the left side, use the chain rule: $\frac{d}{dx}\left[e^{xy}\right] = e^{xy}\cdot \frac{d}{dx}[xy]$, and since $xy$ is a product, $\frac{d}{dx}[xy] = y + x\frac{dy}{dx}$. The right side differentiates to $1 - \frac{dy}{dx}$. So the equation becomes $e^{xy}\left(y + x\frac{dy}{dx}\right) = 1 - \frac{dy}{dx}.$

Expand the left side: $y e^{xy} + x e^{xy}\frac{dy}{dx} = 1 - \frac{dy}{dx}$. Move all $\frac{dy}{dx}$ terms to one side: $x e^{xy}\frac{dy}{dx} + \frac{dy}{dx} = 1 - y e^{xy}$. Factor out $\frac{dy}{dx}$ and solve: $\frac{dy}{dx} = \frac{1 - y e^{xy}}{x e^{xy} + 1}.$