The maximum value of $\left(\frac{1}{x}\right)^x$ is:

$e^{1/e}$

The correct answer is Option (3) → $e^{1/e}$ ##

Let $y = \left(\frac{1}{x}\right)^x$

$\Rightarrow \log y = x \cdot \log \frac{1}{x}$

$∴\frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{1/x} \left(-\frac{1}{x^2}\right) + \log \frac{1}{x} \cdot 1$

$= -1 + \log \frac{1}{x}$

$\frac{dy}{dx} = \left(\log \frac{1}{x} - 1\right) \left(\frac{1}{x}\right)^x$

Now, $\frac{dy}{dx} = 0$

$\Rightarrow \log \frac{1}{x} = 1 = \log e$

$\Rightarrow \frac{1}{x} = e$

$\Rightarrow x = \frac{1}{e}$

Hence, the maximum value of $f\left(\frac{1}{e}\right) = (e)^{1/e}$.