1. **State the problem:** Given functions $g(x) = \frac{1-x}{x}$ and the composition $g \circ f(x) = \frac{1}{x+1}$, find the inverse of the composition, i.e., $(g \circ f)^{-1}$. 2. **Recall the definition of inverse function:** For a function $h(x)$, its inverse $h^{-1}(x)$ satisfies $h(h^{-1}(x)) = x$. 3. **Set $y = (g \circ f)(x) = \frac{1}{x+1}$ and solve for $x$ in terms of $y$ to find $(g \circ f)^{-1}(y)$:** $$y = \frac{1}{x+1}$$ 4. Multiply both sides by $x+1$: $$y(x+1) = 1$$ 5. Distribute $y$: $$yx + y = 1$$ 6. Isolate $x$: $$yx = 1 - y$$ 7. Divide both sides by $y$ (assuming $y \neq 0$): $$x = \frac{1 - y}{y}$$ 8. Using the cancellation notation for division: $$x = \frac{\cancel{1 - y}}{\cancel{y}}$$ 9. Therefore, the inverse function is: $$(g \circ f)^{-1}(y) = \frac{1 - y}{y}$$ 10. **Final answer:** $$(g \circ f)^{-1}(x) = \frac{1 - x}{x}$$