The maximum value of $(\frac{1}{x})^x$ for $x > 0$ is

$e^{1/e}$

The correct answer is Option (2) → $e^{1/e}$

Let $f(x)=x^{1/x}$ for $x>0$.

$\ln f(x)=\frac{\ln x}{x}$

$\frac{d}{dx}\!\left(\frac{\ln x}{x}\right)=\frac{1-\ln x}{x^{2}}=0\;\Rightarrow\;\ln x=1\;\Rightarrow\;x=e$.

At $x=e$, $f(e)=e^{1/e}$ and this is a maximum (since $1-\ln x$ changes sign from $+$ to $-$).

Maximum value: $e^{1/e}$.