Answer
$$ f'(x)= e^{ x^2\ln x}(2x\ln x+x).$$
Work Step by Step
Recall that $(e^x)'=e^x $, $(\ln x)'=\dfrac{1}{x}$.
We have
$$ f(x)=x^{x^2}=e^{ x^2\ln x}.$$
Now taking the derivative, we get
$$ f'(x)= e^{ x^2\ln x}(x^2\ln x)'=e^{ x^2\ln x}(2x\ln x+x).$$
