If $xy = e^{(x-y)}$, then $\frac{dy}{dx}$ is equal to:

Given the equation $xy = e^{(x-y)}$

Using implicit differentiation with respect to $x$, treating $y$ as a function of $x$.

The left side, using the product rule:

$\frac{d}{dx}(xy) = x \cdot \frac{dy}{dx} + y \cdot 1$

$\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$

The right side, using the chain rule:

$\frac{d}{dx}(e^{(x-y)}) = e^{(x-y)} \cdot \frac{d}{dx}(x-y)$

$\frac{d}{dx}(e^{(x-y)}) = e^{(x-y)} \cdot (1 - \frac{dy}{dx})$

$\frac{d}{dx}(e^{(x-y)}) = e^{(x-y)} - e^{(x-y)} \cdot \frac{dy}{dx}$

Setting both sides equal:

$x\frac{dy}{dx} + y = e^{(x-y)} - e^{(x-y)} \cdot \frac{dy}{dx}$

Collecting all terms with $\frac{dy}{dx}$ on one side:

$x\frac{dy}{dx} + e^{(x-y)} \cdot \frac{dy}{dx} = e^{(x-y)} - y$

Factoring out $\frac{dy}{dx}$:

$\frac{dy}{dx}(x + e^{(x-y)}) = e^{(x-y)} - y$

$\frac{dy}{dx} = \dfrac{e^{(x-y)} - y}{x + e^{(x-y)}}$