Explanation: Let $\mathit{y}={\left(\frac{1}{\mathit{x}}\right)}^{\mathit{x}}$
Then, $\log \mathit{y}=\mathit{x}\log \left(\frac{1}{\mathit{x}}\right)=-\mathit{x}\log \mathit{x}$
Differentiating both sides w.r.t. $\mathit{x}$
$\therefore \frac{1}{\mathit{y}}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-\left[\mathit{x}\cdot \frac{1}{\mathit{x}}+\log \mathit{x}\right]$
$=-\left(1+\log \mathit{x}\right)\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(\mathit{i}\right)$
On differentiating again eq. (ii), we get
$\frac{1}{\mathit{y}}\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}-\frac{1}{{\mathit{y}}^2}{\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)}^2=\frac{-1}{\mathit{x}}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(\mathit{i}\mathit{i}\right)$
From eq. (ii), we get
$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=-\mathit{y}\left(1+\log \mathit{x}\right)\cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(\mathit{i}\mathit{i}\mathit{i}\right)$
$=-{\left(\frac{1}{\mathit{x}}\right)}^{\mathit{x}}\left(1+\log \mathit{x}\right)$
For maximum or minimum values of $\mathit{y}$, put $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=0$
Therefore, ${\left(\frac{1}{\mathit{x}}\right)}^{\mathit{x}}\left(1+\log \mathit{x}\right)=0$
However, ${\left(\frac{1}{\mathit{x}}\right)}^{\mathit{x}}\ne 0$ for any value of $\mathit{x}$. Therefore
$1+\log \mathit{x}=0$
$\Rightarrow \log \mathit{x}=-1\Rightarrow \mathit{x}={\mathit{e}}^{-1}\Rightarrow \mathit{x}=\frac{1}{\mathit{e}}$
When $\mathit{x}=\frac{1}{\mathit{e}}$, from eq. (iii)
$\frac{1}{\mathit{y}}\frac{{\mathit{d}}^2\mathit{y}}{\mathit{d}{\mathit{x}}^2}-0=-\mathit{e}$

Hence, $\mathit{y}$ is maximum when $\mathit{x}=\frac{1}{\mathit{e}}$ and maximum value of $\mathit{y}={\mathit{e}}^{1∕\mathit{e}}$