The maximum value of ${\left(\frac{1}{x}\right)}^{2x^2},$ is

Let $f\left(x\right)={\left(\frac{1}{x}\right)}^{2x^2}\cdot$Clearly, $f\left(x\right)$ is defined for all $x>0.$

Now,

$\begin{array}{l}f\left(x\right)=e^{-2x^2{\log }_{e}x}\\ \Rightarrow f^{\mathrm{\prime }}\left(x\right)={\left(\frac{1}{x}\right)}^{2x^2}\left\{-4x{\log }_{e}x-2x\right\}\\ \Rightarrow f^{\mathrm{\prime }}\left(x\right)=-{\left(\frac{1}{x}\right)}^{2x^2}2x\left(2{\log }_{e}x+1\right)\end{array}$

For local maximum, we must have 

$f^{\mathrm{\prime }}\left(x\right)=0\Rightarrow 2x\left(2{\log }_{e}x+1\right)=0\Rightarrow x=e^{-1/2}[\because x>0]$

It can be checked that $f^{\prime \prime }\left(e^{-1/2}\right)<0.$

Therefore, $f\left(x\right)$ attains maximum value at $x=e^{-1/2}$. The maximum value of $f\left(x\right)$ is given by

$f\left(e^{-1/2}\right)={\left(e^{1/2}\right)}^{2/e}=e^{1/e}$