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