The expression inside \(tan^{-1}\) is of the form \( \frac{u}{v} \) where \( u = 3x-x^3 \) and \( v = 1-3x^2 \).
Using the derivative rule for \(tan^{-1}(x)\), which is \(\frac{1}{1+x^2}\), and chain rule, the derivative is \(\frac{1}{1+\left(\frac{3x-x^3}{1-3x^2}\right)^2} \times \frac{d}{dx}(\frac{3x-x^3}{1-3x^2})\).
Upon differentiating and simplifying, we find it matches option \(I\): \(\frac{3}{1+x^2}\).
The given function resembles a standard transformation for trigonometric functions.
The derivative of \(cos^{-1}(x)\) is \(-\frac{1}{\sqrt{1-x^2}}\), but here simplify the expression using trigonometric identities, leading us to \(-\frac{2}{1+x^2}\).