1. **Problem:** Given $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2t}{1 + t^2}$, prove that $$\frac{dy}{dx} + \frac{x}{y} = 0.$$\n\n2. **Find derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$:**\n$$\frac{dx}{dt} = \frac{d}{dt} \left( \frac{1 - t^2}{1 + t^2} \right) = \frac{(0 - 2t)(1 + t^2) - (1 - t^2)(2t)}{(1 + t^2)^2} = \frac{-2t(1 + t^2) - 2t(1 - t^2)}{(1 + t^2)^2}.$$\nSimplify numerator:\n$$-2t(1 + t^2) - 2t(1 - t^2) = -2t - 2t^3 - 2t + 2t^3 = -4t.$$\nSo, $$\frac{dx}{dt} = \frac{-4t}{(1 + t^2)^2}.$$\n\nSimilarly,\n$$\frac{dy}{dt} = \frac{d}{dt} \left( \frac{2t}{1 + t^2} \right) = \frac{2(1 + t^2) - 2t(2t)}{(1 + t^2)^2} = \frac{2 + 2t^2 - 4t^2}{(1 + t^2)^2} = \frac{2 - 2t^2}{(1 + t^2)^2} = \frac{2(1 - t^2)}{(1 + t^2)^2}.$$\n\n3. **Find $\frac{dy}{dx}$:**\n$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{2(1 - t^2)}{(1 + t^2)^2}}{\frac{-4t}{(1 + t^2)^2}} = \frac{2(1 - t^2)}{-4t} = -\frac{1 - t^2}{2t}.$$\n\n4. **Calculate $\frac{x}{y}$:**\n$$\frac{x}{y} = \frac{\frac{1 - t^2}{1 + t^2}}{\frac{2t}{1 + t^2}} = \frac{1 - t^2}{1 + t^2} \times \frac{1 + t^2}{2t} = \frac{1 - t^2}{2t}.$$\n\n5. **Sum $\frac{dy}{dx} + \frac{x}{y}$:**\n$$-\frac{1 - t^2}{2t} + \frac{1 - t^2}{2t} = 0.$$\n\n**Hence proved:** $$\frac{dy}{dx} + \frac{x}{y} = 0.$$