1. **State the problem:** Find the inverse function $f^{-1}(x)$ for the function $$y = -\frac{10}{1 + x^2}, \quad x \geq 0.$$\n\n2. **Rewrite the function:** We have $$y = -\frac{10}{1 + x^2}.$$\n\n3. **Solve for $x$ in terms of $y$: **\nStart by isolating the denominator:\n$$y = -\frac{10}{1 + x^2} \implies y(1 + x^2) = -10.$$\n\n4. **Distribute $y$: **\n$$y + yx^2 = -10.$$\n\n5. **Isolate the $x^2$ term: **\n$$yx^2 = -10 - y.$$\n\n6. **Divide both sides by $y$: **\n$$x^2 = \frac{-10 - y}{y} = \frac{-10}{y} - 1.$$\nShow the cancellation step explicitly:\n$$x^2 = \frac{\cancel{y}(-10/y - 1)}{\cancel{y}} = -\frac{10}{y} - 1.$$\n\n7. **Take the square root (considering $x \geq 0$): **\n$$x = \sqrt{-\frac{10}{y} - 1}.$$\n\n8. **Rewrite the inverse function:**\n$$f^{-1}(x) = \sqrt{-\frac{10}{x} - 1}.$$\n\n9. **Determine the domain of $f^{-1}$:**\nSince the original function outputs values $y$ such that $$y = -\frac{10}{1 + x^2},$$ with $x \geq 0$, the denominator $1 + x^2 \geq 1$, so $$y \in (-10, 0].$$\nBecause $y$ is negative and approaches 0 from below, the range is $(-10, 0]$.\n\n10. **Rewrite the inverse function domain:**\nThe inverse function's domain is the original function's range, so $$x \in (-10, 0].$$\n\n11. **Simplify the inverse function expression:**\nRewrite inside the square root:\n$$-\frac{10}{x} - 1 = \frac{-10 - x}{x}.$$\nSince $x$ is negative in the domain $(-10, 0]$, the expression inside the root is positive.\n\n**Final answer:**\n$$f^{-1}(x) = \sqrt{-\frac{10}{x} - 1}, \quad x \in (-10, 0].$$