The parametric equations provided are:
\[
x = \frac{1 - t^2}{1 + t^2}, \quad y = \frac{2t}{1 + t^2}.
\]
Step 1: Perform a trigonometric substitution.
Let \( t = \tan\theta \). Substituting this into the expressions for \( x \) and \( y \) yields:
\[
x = \cos 2\theta, \quad y = \sin 2\theta.
\]
Step 2: Differentiate both \( x \) and \( y \) with respect to \( \theta \).
The derivatives are:
\[
\frac{dx}{d\theta} = -2\sin 2\theta, \quad \frac{dy}{d\theta} = 2\cos 2\theta.
\]
Calculate \( \frac{dy}{dx} \) using the chain rule:
\[
\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2\cos 2\theta}{-2\sin 2\theta} = -\cot 2\theta.
\]
Step 3: Simplify the resulting expression.
Utilizing trigonometric identities, we obtain:
\[
\frac{dy}{dx} = -\frac{x}{y}.
\]
The solution is \( \boxed{(3)} \).