If the maximum value of the function $f(x) = \frac{2\log_e x}{x}$, $x > 0$ occurs at $x = e$, then $e^3 f''(e)$ is equal to

Given $f(x) = \frac{2\log_e x}{x}$ where $x > 0$, and the maximum value occurs at $x = e$.

For the quotient $\frac{u}{v}$ where $u = 2\ln x$ and $v = x$:

$u' = \frac{2}{x}$ and $v' = 1$

$f'(x) = \frac{x \cdot \frac{2}{x} - 2\ln x \cdot 1}{x^2}$

$f'(x) = \frac{2 - 2\ln x}{x^2}$

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For $f'(x) = \frac{2 - 2\ln x}{x^2}$ where $u = 2 - 2\ln x$ and $v = x^2$:

$u' = -\frac{2}{x}$ and $v' = 2x$

$f''(x) = \frac{x^2 \cdot \left(-\frac{2}{x}\right) - (2 - 2\ln x) \cdot 2x}{(x^2)^2}$

$f''(x) = \frac{-2x - 4x + 4x\ln x}{x^4}$

$f''(x) = \frac{-6x + 4x\ln x}{x^4}$

$f''(x) = \frac{x(-6 + 4\ln x)}{x^4}$

$f''(x) = \frac{-6 + 4\ln x}{x^3}$

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At $x = e$:

$f''(e) = \frac{-6 + 4\ln e}{e^3}$

$f''(e) = \frac{-6 + 4(1)}{e^3}$

$f''(e) = \frac{-2}{e^3}$

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$e^3 f''(e) = e^3 \times \frac{-2}{e^3}$

$e^3 f''(e) = -2$

Therefore, $e^3 f''(e) = -2$