The function $f\left(x\right)=x^{1/x}$ is:
Statement I : increasing in $(1,e)$
Statement II : decreasing in $(e,\infty )$ which of the above statements are true 

Let  $y=x^{1/x}$ 
Taking log on both sides, we get 
          $\log y=\frac{1}{x}\log x$
$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{x^2}-\frac{\log x}{x^2}=\frac{1-\log x}{x^2}$ 
 $\Rightarrow \frac{dy}{dx}=x^{1/x}\left(\frac{1-\log x}{x^2}\right)$
Now, $x^{1/x}>0$ for all positive value of  $x$
And $\frac{1-\log x}{x^2}>0in(1,e)$ 
And $\frac{1-\log x}{x^2}<0$ in  $(e,\infty )$
$\therefore f\left(x\right)$ is increasing in $(1,e)$  and decreasing in $(e,\infty )$ .