Answer
$$g'\left( x \right) = \frac{{2x - {x^2}}}{{{e^x}}}$$
Work Step by Step
$$\eqalign{
& g\left( x \right) = \frac{{{x^2}}}{{{e^x}}} \cr
& {\text{Differentiate}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{{{e^x}}}} \right] \cr
& {\text{Use the quotient rule}} \cr
& g'\left( x \right) = \frac{{{e^x}\left( {{x^2}} \right)' - {x^2}\left( {{e^x}} \right)'}}{{{{\left( {{e^x}} \right)}^2}}} \cr
& g'\left( x \right) = \frac{{{e^x}\left( {2x} \right) - {x^2}\left( {{e^x}} \right)}}{{{e^{2x}}}} \cr
& {\text{Simplify}} \cr
& g'\left( x \right) = \frac{{2x - {x^2}}}{{{e^x}}} \cr} $$
