1. Given function

   $f(x)=\left(\frac{2}{x}\right)^{x^2}, \qquad x>0$

   We need the local maximum value of this function.

2. Take logarithm

   Let $y=f(x)=\left(\frac{2}{x}\right)^{x^2}$ Taking natural log, $\ln y = x^2 \ln\left(\frac{2}{x}\right)=x^2(\ln 2-\ln x)$

   Define $\phi(x)=\ln y=x^2(\ln 2-\ln x)$

   Since $\ln y$ is an increasing function of $y$, maxima/minima of $y$ occur at the same critical points as those of $\phi(x)$.

3. Differentiate

   $\phi'(x)=\frac{d}{dx}\left[x^2(\ln 2-\ln x)\right]$

   Using product rule, $\phi'(x)=2x(\ln 2-\ln x)+x^2\left(-\frac{1}{x}\right)$ $\phi'(x)=2x\ln\left(\frac{2}{x}\right)-x$ $\phi'(x)=x\left(2\ln\left(\frac{2}{x}\right)-1\right)$

4. Find critical points

   Since $x>0$, we set $2\ln\left(\frac{2}{x}\right)-1=0$ $2\ln\left(\frac{2}{x}\right)=1$ $\ln\left(\frac{2}{x}\right)=\frac12$ $\frac{2}{x}=e^{1/2}=\sqrt e$ $x=\frac{2}{\sqrt e}$

5. Check whether it is a maximum

   Consider $\phi'(x)=x\left(2\ln\left(\frac{2}{x}\right)-1\right)$

   • For $x<\frac{2}{\sqrt e}$, we have $\ln(2/x)>1/2$, so $\phi'(x)>0$.
   • For $x>\frac{2}{\sqrt e}$, we have $\ln(2/x)<1/2$, so $\phi'(x)<0$.

   Therefore, the function increases and then decreases, so $x=\frac{2}{\sqrt e}$ gives a local maximum.

6. Compute the maximum value

   $f\left(\frac{2}{\sqrt e}\right)=\left(\frac{2}{2/\sqrt e}\right)^{\left(2/\sqrt e\right)^2}$ $=\left(\sqrt e\right)^{4/e}$ $=(e^{1/2})^{4/e}=e^{2/e}$

7. Match with options

   $e^{2/e}$ corresponds to Option C.

Final Answer: Option C