If $e^x + e^y = e^{x+y}$, then $\frac{dy}{dx}$ =

Given $e^x + e^y = e^{x+y}$

Differentiating both sides with respect to $x$:

$e^x + e^y \cdot \dfrac{dy}{dx} = e^{x+y} \cdot \left(1 + \dfrac{dy}{dx}\right)$

$e^x + e^y \cdot \dfrac{dy}{dx} = e^{x+y} + e^{x+y} \cdot \dfrac{dy}{dx}$

$e^y \cdot \dfrac{dy}{dx} - e^{x+y} \cdot \dfrac{dy}{dx} = e^{x+y} - e^x$

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

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

$\dfrac{dy}{dx} = \dfrac{e^x(e^y - 1)}{-e^y(e^x - 1)}$

From the original equation $e^x + e^y = e^x \cdot e^y$:

$e^y(e^x - 1) = e^x \quad \cdots (i)$

$e^x(e^y - 1) = e^y \quad \cdots (ii)$

Substituting $(i)$ and $(ii)$:

$\dfrac{dy}{dx} = \dfrac{e^y}{-e^x}$

$= -e^{\,y-x}$

$\boxed{\dfrac{dy}{dx} = -e^{\,y-x}}$