Answer
$g'\left( x \right) = \left( {x + 2x\ln x} \right){x^{{x^2}}}$
Work Step by Step
$$\eqalign{
& g\left( x \right) = {e^{{x^2}\ln x}} \cr
& {\text{Differentiate}} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{{x^2}\ln x}}} \right] \cr
& {\text{Use the formula }}\frac{d}{{dx}}\left[ {{e^x}} \right]{\text{ and the chain rule}}{\text{, then}} \cr
& \frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}},{\text{ let }}u = {x^2}\ln x \cr
& g'\left( x \right) = {e^{{x^2}\ln x}}\underbrace {\frac{d}{{dx}}\left[ {{x^2}\ln x} \right]}_{{\text{use product rule}}} \cr
& g'\left( x \right) = {e^{{x^2}\ln x}}\left( {{x^2}\frac{d}{{dx}}\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\left[ {{x^2}} \right]} \right) \cr
& {\text{Computing derivatives}} \cr
& g'\left( x \right) = {e^{{x^2}\ln x}}\left( {{x^2}\left( {\frac{1}{x}} \right) + \ln x\left( {2x} \right)} \right) \cr
& g'\left( x \right) = \left( {x + 2x\ln x} \right){e^{{x^2}\ln x}} \cr
& g'\left( x \right) = \left( {x + 2x\ln x} \right){(e^{\ln x})^{x^2}} \cr
& g'\left( x \right) = \left( {x + 2x\ln x} \right)x^{x^2} \cr}$$
