Let $\mathrm{y}=\left(\frac{1}{\mathrm{x}}\right)^{\mathrm{x}} \Rightarrow \mathrm{y}=\mathrm{x}^{-\mathrm{x}}$
Then $\log y=-x \log x$
$\therefore \frac{1}{y} \frac{d y}{d x}=-(1+\log x)$
or $\frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{y}(1+\log \mathrm{x})$
$\& \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\mathrm{y} \cdot \frac{1}{\mathrm{x}}-(1+\log \mathrm{x}) \cdot \frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}$
$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\left(\frac{1}{\mathrm{x}}\right)^{\mathrm{x}+1}-(1+\log \mathrm{x}) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$
Now, $\frac{\mathrm{dy}}{\mathrm{dx}}=0 \Rightarrow 1+\log \mathrm{x}=0$
$\Rightarrow \log x=-1=-\log \mathrm{e}=\log \left(\frac{1}{\mathrm{e}}\right)$
$\Rightarrow \mathrm{x}=\left(\frac{1}{\mathrm{e}}\right)$
Also, $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$ at $\mathrm{x}=\left(\frac{1}{\mathrm{e}}\right)$ is $-\mathrm{e}^{1+\frac{1}{\mathrm{e}}} < 0$
$\left[\because \frac{\mathrm{dy}}{\mathrm{dx}}=0\right]$
So, $x=\left(\frac{1}{e}\right)$ is a point of local maxima.
$\therefore$ Maximum value
$=\left(\right.$ value of $\mathrm{y}$ when $\left.\mathrm{x}=\frac{1}{\mathrm{e}}\right)=\mathrm{e}^{\frac{1}{\mathrm{e}}}$