The function $f(x)=x^x$ decreases on the interval.

(0, 1/e)

$f(x)=x^x$ (Domain is x > 0)

$y=x^x,\log y = x\log x$

Diff. wrt x,

$\frac{1}{y}\frac{dy}{dx}=x×\frac{1}{x}+\log x⇒\frac{dy}{dx}=x^x(1+\log x)<0$

When 1 + log x < 0 ⇒ log x < -1 $⇒ x < \frac{1}{e}$

As domain of f(x) is x > 0

f(x) decreases when $0<x<\frac{1}{e}\,\,x∈(0,\frac{1}{e})$