Let \( y=\left(\frac{1}{x}\right)^{x} \)
Taking log on both the sides, we get
\[
\begin{array}{l}
\log y=x \log \left(\frac{1}{x}\right)=-x \log x \\
\log y=-x \log x \rightarrow(1)
\end{array}
\]
Now, differentiating Eq. (1) with respect to \( x \), we get
\[
\begin{array}{l}
\frac{1}{y} \frac{d y}{d x}=-\left(x \cdot \frac{1}{x}+\log x\right) \\
\Rightarrow \frac{1}{y} \frac{d y}{d x}=-(1+\log x) \\
\Rightarrow \frac{d y}{d x}=-y(1+\log x) \rightarrow(2) \\
\Rightarrow \frac{d y}{d x}=-\left(\frac{1}{x}\right)^{x}(1+\log x)
\end{array}
\]
\[
\begin{array}{l}
\text { Now, } \frac{d y}{d x}=0 \\
-y(1+\log x)=0 \\
\Rightarrow \log x=-1 \Rightarrow x=e^{-1}
\end{array}
\]
Now, differentiating Eq. (2) with respect to \( x \), we get
\[
\begin{array}{l}
\frac{d^{2} y}{d x^{2}}=-(1+\log x) \frac{d y}{d x}-\frac{y}{x} \\
\frac{d^{2} y}{d x^{2}}=y(1+\log x)^{2}-\frac{y}{x} \\
\frac{d^{2} y}{d x^{2}}=\left(\frac{1}{x}\right)^{x}(1+\log x)^{2}-\left(\frac{1}{x}\right)^{x} \frac{1}{x} \\
\text { At } x=e^{-1}, \text { we get }
\end{array}
\]