To solve this question, let's analyze each statement individually and apply necessary mathematical reasoning to ascertain their accuracy.

Statement-I Analysis: The function given is \(f(x) = \frac{x^3 + 128}{x}\).

To determine whether this function has a minimum at \(x = 4\), we need to find its critical points by differentiating it. We start by calculating the first derivative of \(f(x)\):

\(f(x) = \frac{x^3 + 128}{x} = x^2 + \frac{128}{x}\)

\(f'(x) = \frac{d}{dx}(x^2 + \frac{128}{x}) = 2x - \frac{128}{x^2}\)

Set \(f'(x) = 0\) to find critical points:

\(2x - \frac{128}{x^2} = 0 \Rightarrow 2x^3 = 128 \Rightarrow x^3 = 64 \Rightarrow x = 4\)

Next, calculate the second derivative to confirm it's a minimum:

\(f''(x) = \frac{d}{dx}(2x + \frac{128}{x^2}) = 2 + \frac{256}{x^3}\)

At \(x = 4\), \(f''(4) = 2 + \frac{256}{64} = 2 + 4 = 6\), which is positive, indicating a local minimum at \(x = 4\). So, Statement-I is correct.

Statement-II Analysis: We need to assess the sign change in \(f'(x)\) around \(x=4\).

Consider \(f'(x) = 2x - \frac{128}{x^2}\).

- For values just less than 4, say \(x = 3.9\), \(f'(3.9) = 2(3.9) - \frac{128}{(3.9)^2}\\), which will be positive.

- For values just more than 4, say \(x = 4.1\), \(f'(4.1) = 2(4.1) - \frac{128}{(4.1)^2}\\), which will also remain positive.

This implies that \(f'(x)\) does not change from positive to negative as it crosses \(x = 4\). Thus, Statement-II is incorrect.

Conclusion: Analyzing both statements, Statement-I is correct in identifying a minimum value for the function at \(x=4\), but Statement-II is incorrect about the behavior of \(f'(x)\) as it does not change sign at \(x=4\).

Therefore, the correct answer is: Statement-I is correct but Statement-II is not correct.