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

Clearly, f(x) is defined for x > 0

Now, $f(x)=x^x⇒f'(x)=x^x(1+\log x)$

For to be decreasing, we must have

$f'(x)<0⇒x^x(1+\log x)<0$

$⇒1+\log x<0⇒\log x< -1 ⇒ x<e^{-1}$

So, f(x) is decreasing on $(0,\frac{1}{e})$