xx has a stationary point at

$y=x^x \Rightarrow \log y=x \log x$

$\Rightarrow \frac{1}{y} \frac{d y}{d x}=1+\log x \Rightarrow \frac{d y}{d x}=y(1+\log x)$

$\frac{d y}{d x}=0 \Rightarrow 1+\log x=0 \Rightarrow x=e^{-1}$