Given $f(x) = y = x^x$
$\Rightarrow \, \log \, y = x \,\log \, x$
Differentiating, we get
$\frac{1}{y} . \frac{dy}{dx} = \frac{x}{x} + \log \,x$
$ \Rightarrow \frac{dy}{dx} = y \left[1+\log x\right] $
At the stationary point, $ \frac{dy}{dx} = 0 $
$ \Rightarrow x^{x} \left[1+ \log x \right] = 0$
$ \Rightarrow \log_{e} x = -1$
$ \Rightarrow x = \frac{1}{e} $