1. Given function

We have $f(x)=\left(\frac{1}{x}\right)^{2x}=x^{-2x}, \qquad x>0.$

We are told that this function attains its maximum at $x=\frac{1}{e}.$

2. Take logarithm to analyze the function

Let $y=f(x)=x^{-2x}.$ Then $\ln y=\ln\left(x^{-2x}\right)=-2x\ln x.$

So maximizing $f(x)$ is equivalent to maximizing $\phi(x)=-2x\ln x.$

Differentiate: $\phi'(x)=-2(\ln x+1).$

Set $\phi'(x)=0$: $-2(\ln x+1)=0$ $\ln x=-1$ $x=\frac{1}{e}.$

Also, $\phi''(x)=-\frac{2}{x}<0 \quad (x>0),$ so this critical point is indeed a maximum.

Thus the given statement is correct.

3. Use the maximum property

Since $f(x)$ is maximum at $x=\frac{1}{e}$, for every $x>0$, $f\left(\frac{1}{e}\right)>f(x)$ for $x\neq \frac{1}{e}$.

Now, $f\left(\frac{1}{e}\right)=\left(e\right)^{2/e}=e^{2/e}.$

Choose a suitable value of $x$ to compare $e$ and $\pi$. Take $x=\frac{1}{\pi}.$ Then $f\left(\frac{1}{\pi}\right)=\left(\pi\right)^{2/\pi}=\pi^{2/\pi}.$

Since $\frac{1}{\pi}\neq \frac{1}{e}$, we get $e^{2/e}>\pi^{2/\pi}.$

4. Convert this inequality

Raise both sides to the power $\dfrac{e\pi}{2}>0$:

$\left(e^{2/e}\right)^{e\pi/2}>\left(\pi^{2/\pi}\right)^{e\pi/2}.$

Simplifying, $e^{\pi}>\pi^{e}.$

This matches option D.

5. Check options

• A: $e^\pi<\pi^e$ — false, opposite of what we proved.
• B: $e^{2\pi}<(2\pi)^e$ — not implied; in fact this is false.
• C: $(2e)^\pi>\pi^{2e}$ — not generally following from the given result.
• D: $e^\pi>\pi^e$ — true.

6. Final answer

The correct option is $\boxed{D}.$