Answer
$$\frac{{dy}}{{dx}} = - {e^{ - x}}$$
Work Step by Step
$$\eqalign{
& {e^x}y = 1 \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \underbrace {\frac{d}{{dx}}\left[ {{e^x}y} \right]}_{{\text{Product rule}}} = \frac{d}{{dx}}\left[ 1 \right] \cr
& {e^x}\frac{d}{{dx}}\left[ y \right] + y\frac{d}{{dx}}\left[ {{e^x}} \right] = 0 \cr
& {e^x}\frac{{dy}}{{dx}} + {e^x}y = 0 \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& {e^x}\frac{{dy}}{{dx}} = - {e^x}y \cr
& \frac{{dy}}{{dx}} = - \frac{{{e^x}y}}{{{e^x}}} \cr
& \frac{{dy}}{{dx}} = - y \cr
& \cr
& {\text{Solving for }}y \cr
& {e^x}y = 1 \cr
& y = {e^{ - x}} \cr
& {\text{Differentiate }} \cr
& y = - {e^{ - x}} \cr
& \cr
& {\text{Substituting }}y = {e^{ - x}}{\text{ into }}\frac{{dy}}{{dx}} = - y \cr
& \frac{{dy}}{{dx}} = - {e^{ - x}} \cr} $$
