To determine if \(f(x) = \frac{1}{x^3} - x^3\) is a decreasing function, we need to find its derivative \(f'(x)\) and check if it is negative for \(x > 0\).

First, differentiate \(f(x)\):

\(f'(x) = \frac{d}{dx} \left( \frac{1}{x^3} \right) - \frac{d}{dx} (x^3)\).

Using the power rule, \(\frac{d}{dx} \left( x^n \right) = nx^{n-1}\), we find:

\(\frac{d}{dx} \left( \frac{1}{x^3} \right) = \frac{d}{dx} (x^{-3}) = -3x^{-4}\).

\(\frac{d}{dx} (x^3) = 3x^2\).

Thus, \(f'(x) = -3x^{-4} - 3x^2\).

Simplifying, \(f'(x) = -\frac{3}{x^4} - 3x^2\).

For \(x > 0\), both terms \(-\frac{3}{x^4}\) and \(-3x^2\) are negative, so \(f'(x) < 0\).

Therefore, \(f(x)\) is a decreasing function for \(x > 0\).