Answer
$$\eqalign{
& g'\left( x \right) = 2{x^{\ln x - 1}}\ln x \cr
& g'\left( e \right) = 2 \cr} $$
Work Step by Step
$$\eqalign{
& g\left( x \right) = {x^{\ln x}};\,\,\,\,\,\,a = e \cr
& {\text{Use }}{b^x} = {e^{x\ln b}},{\text{ then}} \cr
& g\left( x \right) = {e^{\ln x\ln x}} \cr
& g\left( x \right) = {e^{{{\ln }^2}x}} \cr
& {\text{Differentiating}} \cr
& g'\left( x \right) = {e^{{{\ln }^2}x}}\frac{d}{{dx}}\left[ {{{\ln }^2}x} \right] \cr
& g'\left( x \right) = {e^{{{\ln }^2}x}}\left( {2\ln x} \right)\left( {\frac{1}{x}} \right) \cr
& g'\left( x \right) = \frac{{2{e^{{{\ln }^2}x}}\ln x}}{x} \cr
& g'\left( x \right) = \frac{{2\left( {{x^{\ln x}}} \right)\ln x}}{x} \cr
& g'\left( x \right) = 2{x^{\ln x - 1}}\ln x \cr
& \cr
& {\text{Evaluate the derivative at }}x = e \cr
& g'\left( e \right) = 2{e^{\ln e - 1}}\ln e \cr
& g'\left( e \right) = 2{e^0}\left( 1 \right) \cr
& g'\left( e \right) = 2 \cr} $$
