3. Let $f(x)=\frac{x-1}{x+1}$, and denote $f_{1}(x)=f(x)$. If $f_{n+1}(x)=f\left(f_{n}(x)\right)$, then $f_{2010}(x)=(\quad)$.
(A) $x$
(B) $-\frac{1}{x}$
(C) $\frac{x-1}{x+1}$
(D) $\frac{1+x}{1-x}$
3. B.

From the problem, we can predict that $f_{n}(x)$ is periodic. To find the period, using the substitution method makes the calculation straightforward.
Let $a_{1}=f_{1}(2), a_{n}=f_{n}(2)$. Then
$$
\begin{array}{l}
a_{1}=\frac{1}{3}, a_{2}=\frac{\frac{1}{3}-1}{\frac{1}{3}+1}=-\frac{1}{2}, \\
a_{3}=\frac{-\frac{1}{2}-1}{-\frac{1}{2}+1}=-3, a_{4}=\frac{-3-1}{-3+1}=2, \\
a_{5}=\frac{2-1}{2+1}=\frac{1}{3}, \cdots \cdots
\end{array}
$$

Thus, the period $T=4$ is known.
Therefore, $a_{2010}=a_{2}=-\frac{1}{2}$.
Substituting $x=2$ into each option for verification, we find that the answer is (B).