Problem 1254

(1) Proof by contradiction. If $\frac{1}{4}$ is in the domain of $f(x)$. Then
$$f\left(\frac{1}{4}\right)=f\left(\frac{1}{2} \times \frac{1}{2}\right)=f\left(\frac{1}{2}\right)+f\left(\frac{1}{2}\right)=2$$

This contradicts the range of $f(x)$ being $[-1,1]$. Therefore, $\frac{1}{4}$ is not in the domain.
(2) In $f(x y)=f(x)+f(y)$, let $x=f^{-1}(u), y=f^{-1}(v)$, we get $f\left[f^{-1}(u) \cdot f^{-1}(v)\right]=$ $u+v$, that is
$$f^{-1}(u) \cdot f^{-1}(v)=f^{-1}(u+v)$$

Also, $f^{-1}(1)=\frac{1}{2}$, so the given inequality becomes
$$f^{-1}\left(x+\frac{1}{1-x}\right) \leqslant f^{-1}(1)$$

Since $f(x)$ is a decreasing function, $f^{-1}(u)$ is also decreasing on $[-1,1]$. The given inequality is equivalent to
$$\left\{\begin{array}{l} x+\frac{1}{1-x} \geqslant 1 \\ -1 \leqslant x \leqslant 1 \\ -1 \leqslant \frac{1}{1-x} \leqslant 1 \end{array} \text {, solving gives } x=0 .\right.$$

Therefore, the solution set of the given inequality is $\{0\}$.