1. **State the problem:** Find the inverse function $f^{-1}(x)$ and its derivative given $f(x) = \left(\frac{x-1}{x+1}\right)^3$. 2. **Recall the formula for inverse functions:** To find $f^{-1}(x)$, solve $y = \left(\frac{x-1}{x+1}\right)^3$ for $x$ in terms of $y$. 3. **Undo the cube:** Start by taking the cube root of both sides: $$\sqrt[3]{y} = \frac{x-1}{x+1}$$ 4. **Solve for $x$:** Multiply both sides by $x+1$: $$\sqrt[3]{y}(x+1) = x-1$$ 5. **Distribute:** $$\sqrt[3]{y}x + \sqrt[3]{y} = x - 1$$ 6. **Group $x$ terms on one side:** $$\sqrt[3]{y}x - x = -1 - \sqrt[3]{y}$$ 7. **Factor out $x$:** $$x(\sqrt[3]{y} - 1) = -1 - \sqrt[3]{y}$$ 8. **Divide both sides:** $$x = \frac{-1 - \sqrt[3]{y}}{\sqrt[3]{y} - 1}$$ 9. **Simplify the fraction by multiplying numerator and denominator by $-1$:** $$x = \frac{1 + \sqrt[3]{y}}{1 - \sqrt[3]{y}}$$ 10. **Replace $y$ by $x$ to write the inverse function:** $$f^{-1}(x) = \frac{1 + \sqrt[3]{x}}{1 - \sqrt[3]{x}}$$ 11. **Find the derivative of the inverse function:** Use the quotient rule for $$f^{-1}(x) = \frac{1 + x^{1/3}}{1 - x^{1/3}}$$ 12. **Set numerator and denominator:** $$u = 1 + x^{1/3}, \quad v = 1 - x^{1/3}$$ 13. **Derivatives:** $$u' = \frac{1}{3}x^{-2/3}, \quad v' = -\frac{1}{3}x^{-2/3}$$ 14. **Apply quotient rule:** $$\left(f^{-1}\right)'(x) = \frac{u'v - uv'}{v^2} = \frac{\frac{1}{3}x^{-2/3}(1 - x^{1/3}) - (1 + x^{1/3})(-\frac{1}{3}x^{-2/3})}{(1 - x^{1/3})^2}$$ 15. **Simplify numerator:** $$= \frac{\frac{1}{3}x^{-2/3}(1 - x^{1/3}) + \frac{1}{3}x^{-2/3}(1 + x^{1/3})}{(1 - x^{1/3})^2} = \frac{\frac{1}{3}x^{-2/3}[(1 - x^{1/3}) + (1 + x^{1/3})]}{(1 - x^{1/3})^2}$$ 16. **Simplify inside brackets:** $$= \frac{\frac{1}{3}x^{-2/3}(2)}{(1 - x^{1/3})^2} = \frac{2}{3} \frac{x^{-2/3}}{(1 - x^{1/3})^2}$$ **Final answers:** $$f^{-1}(x) = \frac{1 + \sqrt[3]{x}}{1 - \sqrt[3]{x}}$$ $$\left(f^{-1}\right)'(x) = \frac{2}{3} \frac{x^{-2/3}}{(1 - \sqrt[3]{x})^2}$$