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

Given that $e^x+e^y=e^{x+y}$

Differentiate both sides with respect to $x$

$\begin{array}{rcl}{e}^x+e^y\left(\frac{dy}{dx}\right)& =& e^{x+y}\left(1+\frac{dy}{dx}\right)\\ \frac{dy}{dx}\left(e^y-e^{x+y}\right)& =& e^{x+y}-e^x\\ \frac{dy}{dx}\left(e^y-e^x-e^y\right)& =& e^x+e^y-e^x\\ \frac{dy}{dx}\left(e^x\right)& =& -e^y\\ \frac{dy}{dx}& =& -e^{y-x}\\ & & \end{array}$