If $y=\left(\frac{1}{x}\right)^x$, then value of $e^e\left(\frac{d^2 y}{d x^2}\right)_{x=e}$ is:

$4-\frac{1}{e}$

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

taking log on both sides

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

⇒  $\log y = -x\log x$

differentiating both sides wrt x

so $\frac{1}{x} \frac{dy}{dx} = -\log x - \frac{x}{x}$

so  $\frac{dy}{dx} = y(-\log x -1)$           ......(2)

differentiating both sides wrt x

$\frac{d^2 y}{d x^2}=\frac{d y}{d x} (-\log x - 1) + y(-\frac{1}{x})$           .....(3)

$y]_{x=e}=\left(\frac{1}{e}\right)^e=\frac{1}{e^e}$       .....(4)

$\left.\frac{d y}{d x}\right]_{x=e}  =\frac{1}{e^e}(-\log e-1)$        ........(5)

$=\frac{-2}{e^e}$

so from (3) at x = e

$\left.\frac{d^2 y}{d x^2}\right]_{x=e} =\frac{-2}{e^e}(-\log e-1)+\frac{1}{e^e}(\frac{-1}{e})$

$=\frac{4}{e^e}+\frac{1}{e^e}\left(\frac{-1}{e}\right)$

so $\left.e^e \frac{d^2 y}{d x^2}\right]_{x=e}=4 -\frac{1}{e}$