Find the domain of the function $f(x) = \tan(\arccos(x^2)).$
For $\arccos (x^2)$ to be defined, we must have $-1 \le x^2 \le 1,$ which is satisfied only for $-1 \le x \le 1.$  Then $\arccos (x^2)$ will always return an angle between 0 and $\frac{\pi}{2}.$  Then $\tan (\arccos(x^2))$ is defined, unless $\arccos(x^2) = \frac{\pi}{2}.$  This occurs only when $x = 0.$

Therefore, the domain of $f(x)$ is $\boxed{[-1,0) \cup (0,1]}.$
For $\arccos (x^2)$ to be defined, we must have $-1 \le x^2 \le 1,$ which is satisfied only for $-1 \le x \le 1.$  Then $\arccos (x^2)$ will always return an angle between 0 and $\frac{\pi}{2}.$  Then $\tan (\arccos(x^2))$ is defined, unless $\arccos(x^2) = \frac{\pi}{2}.$  This occurs only when $x = 0.$

Therefore, the domain of $f(x)$ is $\boxed{[-1,0) \cup (0,1]}.$
The final answer is $[-1,0) ∪ (0,1]$. I hope it is correct.