1. **State the problem:** Find the derivative $y'$ of the function $y = \tan^{-1}(x^3)$.\n\n2. **Recall the formula:** The derivative of $\tan^{-1}(u)$ with respect to $x$ is given by $$\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1+u^2} \cdot \frac{du}{dx}.$$\n\n3. **Identify $u$ and find $\frac{du}{dx}$:** Here, $u = x^3$, so $$\frac{du}{dx} = 3x^2.$$\n\n4. **Apply the chain rule:** Using the formula, we get $$y' = \frac{1}{1+(x^3)^2} \cdot 3x^2 = \frac{3x^2}{1+x^6}.$$\n\n5. **Final answer:** $$\boxed{y' = \frac{3x^2}{1+x^6}}.$$