Answer
$$g'\left( x \right) = \frac{1}{{1 + {e^x}}}$$
Work Step by Step
$$\eqalign{
& g\left( x \right) = \ln \frac{{{e^x}}}{{1 + {e^x}}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {\ln \frac{{{e^x}}}{{1 + {e^x}}}} \right] \cr
& g'\left( x \right) = \frac{{1 + {e^x}}}{{{e^x}}}\frac{d}{{dx}}\left[ {\frac{{{e^x}}}{{1 + {e^x}}}} \right] \cr
& {\text{Differentiating}} \cr
& g'\left( x \right) = \frac{{1 + {e^x}}}{{{e^x}}}\left( {\frac{{\left( {1 + {e^x}} \right){e^x} - {e^x}\left( {{e^x}} \right)}}{{{{\left( {1 + {e^x}} \right)}^2}}}} \right) \cr
& {\text{Multiply and simplify}} \cr
& g'\left( x \right) = \frac{{1 + {e^x}}}{{{e^x}}}\left( {\frac{{{e^x} + {e^{2x}} - {e^{2x}}}}{{{{\left( {1 + {e^x}} \right)}^2}}}} \right) \cr
& g'\left( x \right) = \frac{{1 + {e^x}}}{{{e^x}}}\left( {\frac{{{e^x}}}{{{{\left( {1 + {e^x}} \right)}^2}}}} \right) \cr
& g'\left( x \right) = \frac{1}{{1 + {e^x}}} \cr} $$
