Step 1: Spot the two conditions for the function to exist.
We have $f(x) = \sin^{-1}\left(\log_2\left(\frac{x^2}{2}\right)\right)$. First, the log needs a positive input. Second, whatever the log gives must lie inside $[-1, 1]$ for the inverse sine to make sense.
Step 2: Keep the log input positive.
Since $\frac{x^2}{2} \gt 0$ for every $x \ne 0$, the only thing the log rules out is $x = 0$.
Step 3: Write the inverse sine condition.
We need $-1 \le \log_2\left(\frac{x^2}{2}\right) \le 1$.
Step 4: Convert the log bounds into bounds on $\frac{x^2}{2}$.
Raising 2 to each part: $2^{-1} \le \frac{x^2}{2} \le 2^{1}$, that is $\frac{1}{2} \le \frac{x^2}{2} \le 2$.
Step 5: Solve for $x^2$.
Multiply through by 2: $1 \le x^2 \le 4$.
Step 6: Translate $x^2$ into intervals for $x$.
$x^2 \ge 1$ means $x \le -1$ or $x \ge 1$. And $x^2 \le 4$ means $-2 \le x \le 2$. Putting these together gives $-2 \le x \le -1$ or $1 \le x \le 2$.
\[ \boxed{[-2, -1] \cup [1, 2]} \]