xx has a stationary point at

x = e x = $\frac{1}{e}$ x = 1 x = $\sqrt{e}$

x = $\frac{1}{e}$

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