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

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

Now,

$f(x)=x^x \Rightarrow f'(x)=x^x(1+\log x)$

For f(x) to be decreasing, we must have

f'(x) < 0

$\Rightarrow x^x(1+\log x)<0$

$\Rightarrow 1+\log x<0$

$\Rightarrow \log x<-1 \Rightarrow x<e^{-1}$

So, f(x) is decreasing on (0, 1/e).