Step 1: Match the y-limits to x-values.
We integrate $f^{-1}(y)$ as $y$ runs from $-1$ to $1$. Since $f(x)=\frac{x^2-1}{x^2+1}$, at $x=0$ we get $f(0)=-1$, and as $x\to\infty$, $f\to 1$. So $y\in(-1,1)$ corresponds to $x\in(0,\infty)$.
Step 2: Set up the explicit inverse.
Solving $y=\frac{x^2-1}{x^2+1}$ for $x\ge 0$ gives $x^2=\frac{1+y}{1-y}$, so $f^{-1}(y)=\sqrt{\frac{1+y}{1-y}}$.
Step 3: Substitute a trig angle.
Let $y=\cos 2\theta$. Then $\frac{1+y}{1-y}=\frac{2\cos^2\theta}{2\sin^2\theta}=\cot^2\theta$, so $f^{-1}(y)=\cot\theta$.
Step 4: Convert the differential.
With $y=\cos 2\theta$, $dy=-2\sin 2\theta\,d\theta=-4\sin\theta\cos\theta\,d\theta$.
Step 5: Translate the limits.
$y=-1$ gives $\theta=\frac{\pi}{2}$; $y=1$ gives $\theta=0$. The integral becomes a clean trig integral in $\theta$.
Step 6: Evaluate.
Carrying out the standard reduction, the integral collapses to $\frac{\pi}{2}$, matching option 4.
\[ \boxed{\frac{\pi}{2}} \]