Simplify the Limit Relation

The given relation is:

$x = \lim_{n \to \infty} \frac{\sum_{k=1}^{n} \left[ k^2 (f(x))^x \right]}{n^3}$

Let $y = (f(x))^x$. The expression becomes:

$x = \lim_{n \to \infty} \frac{\sum_{k=1}^{n} \left[ k^2 y \right]}{n^3}$

Using the property $[z] \le z < [z] + 1$, we have:

$\sum_{k=1}^{n} k^2 y - n \le \sum_{k=1}^{n} \left[ k^2 y \right] < \sum_{k=1}^{n} k^2 y + n$

Divide by $n^3$ and take the limit $n \to \infty$. The sum $\sum_{k=1}^{n} k^2 y = y \sum_{k=1}^{n} k^2 = y \frac{n(n+1)(2n+1)}{6}$.

The limit of the term $\frac{y \frac{n(n+1)(2n+1)}{6}}{n^3}$ as $n \to \infty$ is $y \cdot \frac{2}{6} = \frac{y}{3}$. The terms $\frac{n}{n^3}$ vanish as $n \to \infty$. By the Squeeze Theorem, the limit is $\frac{y}{3}$.

Thus, the relation simplifies to $x = \frac{y}{3}$.

Solve for the Function $f(x)$

Substitute back $y = (f(x))^x$:

$x = \frac{(f(x))^x}{3}$

Rearrange the terms:

$3x = (f(x))^x$

To isolate $f(x)$, raise both sides to the power of $\frac{1}{x}$:

$(3x)^{\frac{1}{x}} = \left( (f(x))^x \right)^{\frac{1}{x}}$

This gives:

$f(x) = (3x)^{\frac{1}{x}}$

Calculate the Derivative $f'(x)$

Use logarithmic differentiation. Let $y = f(x) = (3x)^{\frac{1}{x}}$.

Take the natural logarithm of both sides:

$\log y = \log \left( (3x)^{\frac{1}{x}} \right)$

Apply logarithm properties:

$\log y = \frac{1}{x} \log(3x)$

Differentiate both sides with respect to $x$ using the quotient rule on the right side:

$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left( \frac{\log(3x)}{x} \right)$

$\frac{1}{y} \frac{dy}{dx} = \frac{(\frac{3}{3x}) \cdot x - \log(3x) \cdot 1}{x^2}$

$\frac{1}{y} \frac{dy}{dx} = \frac{\frac{1}{x} \cdot x - \log(3x)}{x^2}$

$\frac{1}{y} \frac{dy}{dx} = \frac{1 - \log(3x)}{x^2}$

Solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = y \left( \frac{1 - \log(3x)}{x^2} \right)$

Substitute back $y = (3x)^{\frac{1}{x}}$:

$f'(x) = (3x)^{\frac{1}{x}} \left[ \frac{1 - \log(3x)}{x^2} \right]$

This matches Option 3.