Official solution
\[f(x) = {\left( {{2 \over x}} \right)^{{x^2}}}\] ; x > 0
\[\ln f(x) = {x^2}(\ln 2 - \ln x)\]
\[f'(x) = f(x)\{ - x + (\ln 2 - \ln x)2x\} \]
\[f'(x) = \underbrace {f(x)}_ + \,.\,\underbrace x_ + \underbrace {(2\ln 2 - 2\ln x - 1)}_{g(x)}\]
\[g(x) = 2{\ln ^2} - 2\ln x - 1\]
\[ = \ln {4 \over {{x^2}}} - 1 = 0 \Rightarrow x = {2 \over {\sqrt e }}\]
\[LM = {2 \over {\sqrt e }}\]
Local maximum value = \[{\left( {{2 \over {2/\sqrt e }}} \right)^{{4 \over e}}} = {e^{{2 \over e}}}\]