We need the domain of $f(x)=\cos^{-1}\left(\frac{2\sin^{-1}\left(\frac{1}{4x^2-1}\right)}{\pi}\right).$

For this function to be defined, we must ensure:

1. $\sin^{-1}\left(\frac{1}{4x^2-1}\right)$ is defined.
2. The argument of $\cos^{-1}$ lies in $[-1,1]$.

Step 1: Condition for the inverse sine

We need $-1 \le \frac{1}{4x^2-1} \le 1,$ with also $4x^2-1 \ne 0 \quad \Rightarrow \quad x\ne \pm \frac12.$

Now solve $\left|\frac{1}{4x^2-1}\right|\le 1.$ This gives $|4x^2-1|\ge 1.$

So, $4x^2-1\ge 1 \quad \text{or} \quad 4x^2-1\le -1.$

Case 1:

$4x^2-1\ge 1$ $4x^2\ge 2$ $x^2\ge \frac12$ $|x|\ge \frac{1}{\sqrt2}.$

Case 2:

$4x^2-1\le -1$ $4x^2\le 0$ $x=0.$

Thus from the $\sin^{-1}$ condition, $x\in (-\infty,-\tfrac{1}{\sqrt2}]\cup\{0\}\cup[\tfrac{1}{\sqrt2},\infty).$

Step 2: Condition for the inverse cosine

Let $\theta=\sin^{-1}\left(\frac{1}{4x^2-1}\right).$ Then $\theta\in\left[-\frac\pi2,\frac\pi2\right].$ Hence $\frac{2\theta}{\pi}\in[-1,1].$

Therefore the argument of $\cos^{-1}$ is automatically in its allowed domain whenever $\sin^{-1}$ is defined.

So no extra restriction arises from $\cos^{-1}$.

Step 3: Final domain

Hence the domain is $(-\infty,-\tfrac{1}{\sqrt2}]\cup[\tfrac{1}{\sqrt2},\infty)\cup\{0\}.$

This matches Option D.

Step 4: Compare with stored correct answer

Stored correct answer: D

Our derived answer: D

So the answer agrees with the stored correct answer.