Given $f'(x) = 3(x^2 + 2x) - \frac{4}{x^3} + 5$. To find $f(x)$, we integrate $f'(x)$ with respect to $x$. Integrating term by term:
- The integral of $3(x^2 + 2x)$ is $x^3 + 3x^2$.
- The integral of $-\frac{4}{x^3}$ is $\frac{2}{x^2}$.
- The integral of 5 is $5x$.
Thus, $f(x) = x^3 + 3x^2 + \frac{2}{x^2} + 5x + C$.
Given $f(1) = 0$. Substituting $x = 1$:
\[ 0 = 1^3 + 3(1^2) + \frac{2}{1^2} + 5(1) + C = 1 + 3 + 2 + 5 + C. \]
Simplifying, $0 = 11 + C$, so $C = -11$.
Therefore, $f(x) = x^3 + 3x^2 + \frac{2}{x^2} + 5x - 11$.
This corresponds to option (3).