Put \(\tan \theta = x\) Then, \[ y = \tan^{-1} \left( \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right) = \tan^{-1}(\tan 3\theta) = 3\theta \] Now, \[ \frac{dy}{dx} = \frac{dy}{d\theta} \times \frac{d\theta}{dx} = 3 \times \frac{1}{\sec^2 \theta} = \frac{3}{1 + x^2} \]