[Solution] Let $\frac{1-x^{2}}{1+x^{2}}=t$, then $x^{2}=\frac{1-t}{1+t}$.
Thus, the given function becomes $f(t)=\sqrt{\frac{1-t}{1+t}}$.

Therefore,
$$
\begin{aligned}
f\left(\frac{2 x}{1+x^{2}}\right) & =\sqrt{\frac{1-\frac{2 x}{1+x^{2}}}{1+\frac{2 x}{1+x^{2}}}}=\sqrt{\left(\frac{1-x}{1+x}\right)^{2}} \\
& =\left|\frac{1-x}{1+x}\right| \\
& =\frac{1-x}{1+x} \text { or } \frac{x-1}{1+x} .
\end{aligned}
$$

Hence, the correct choice is $(A)$.
7・113 Let $f\left(\frac{1-x^{2}}{1+x^{2}}\right)=x$, then $f\left(\frac{2 x}{1+x^{2}}\right)$ equals
(A) $\frac{1-x}{1+x}$ or $\frac{x-1}{1+x}$.
(B) $\frac{1+x}{1-x}$ or $\frac{1+x}{x-1}$.
(C) $\frac{x-1}{x+1}$.
(D) $\frac{x+1}{x-1}$.
(China Sichuan Province High School Mathematics Competition, 1991)