Answer
$dy = \frac{1}{{{{\left( {1 - {e^x}} \right)}^2}}}dx$
Work Step by Step
$$\eqalign{
& y = \frac{{{e^x}}}{{1 - {e^x}}} \cr
& {\text{Differentiate the function }} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{{e^x}}}{{1 - {e^x}}}} \right] \cr
& {\text{Use the quotient rule for derivatives}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {1 - {e^x}} \right)\frac{d}{{dx}}\left[ {{e^x}} \right] - {e^x}\frac{d}{{dx}}\left[ {1 - {e^x}} \right]}}{{{{\left( {1 - {e^x}} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {1 - {e^x}} \right)\left( {{e^x}} \right) - {e^x}\left( { - {e^x}} \right)}}{{{{\left( {1 - {e^x}} \right)}^2}}} \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{1 - {e^{2x}} + {e^{2x}}}}{{{{\left( {1 - {e^x}} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{{{\left( {1 - {e^x}} \right)}^2}}} \cr
& {\text{Write in the differential form }}dy = f'\left( x \right)dx \cr
& dy = \frac{1}{{{{\left( {1 - {e^x}} \right)}^2}}}dx \cr} $$
