Answer
$$g'\left( x \right) = - \frac{{{e^{ - 2x}}\left( {2x + 1} \right)}}{{{x^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{e^{ - x}}}}{{x{e^x}}} \cr
& f\left( x \right) = \frac{{{e^{ - x}}}}{{{e^x}}}\left( {\frac{1}{x}} \right) \cr
& {\text{Rewrite, recall the property }}\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}} \cr
& f\left( x \right) = {e^{ - x - x}}\left( {\frac{1}{x}} \right) \cr
& f\left( x \right) = \frac{{{e^{ - 2x}}}}{x} \cr
& {\text{Differentiate}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{e^{ - 2x}}}}{x}} \right] \cr
& {\text{By the quotient rule}} \cr
& g'\left( x \right) = \frac{{x\frac{d}{{dx}}\left[ {{e^{ - 2x}}} \right] - {e^{ - 2x}}\frac{d}{{dx}}\left[ x \right]}}{{{{\left( x \right)}^2}}} \cr
& g'\left( x \right) = \frac{{x\left( { - 2{e^{ - 2x}}} \right) - {e^{ - 2x}}\left( 1 \right)}}{{{{\left( x \right)}^2}}} \cr
& g'\left( x \right) = \frac{{ - 2x{e^{ - 2x}} - {e^{ - 2x}}}}{{{x^2}}} \cr
& g'\left( x \right) = - \frac{{{e^{ - 2x}}\left( {2x + 1} \right)}}{{{x^2}}} \cr} $$
