The local maximum value of the function, f x = 2 x x 2 ,   x > 0 , is

$f^{\text{'}}\left(x\right)=0$ for maximum value
Let $y={\left(\frac{2}{x}\right)}^{x^2}$

$\ln y=x^2\ln \frac{2}{x}$

$\frac{1}{y}{y}^{\text{'}}=2x\ln \frac{2}{x}+x^2\frac{1}{\frac{2}{x}}\times \frac{-2}{x^2}$

$y^{\text{'}}=\left(xy\right)\left(2\ln \frac{2}{x}-1\right)$

$y^{\text{'}}={\left(\frac{2}{x}\right)}^{x^2}\left(x\right)\left(2\ln \frac{2}{x}-1\right)$

$2\ln \frac{2}{x}=1$

$\left(\frac{2}{x}\right)=e^{\frac{1}{2}}$

$x=2e^{-\frac{1}{2}}$

Then maximum value will be
$f\left(2e^{-\frac{1}{2}}\right)={\left(\frac{2}{2e^{-\frac{1}{2}}}\right)}^{4e^{-1}}=e^{2e^{-1}}=e^{\frac{2}{e}}$