14. Given the function $f(x)=\frac{x^{3}+3 x}{3 x^{2}+1}$, the sequence $\left\{x_{n}\right\}$ satisfies: $x_{1}=2, x_{n+1}=f\left(x_{n}\right)\left(n \in N^{*}\right)$, and let $b_{n}=\log _{3}\left(\frac{x_{n+1}-1}{x_{n+1}+1}\right) \quad\left(n \in N^{*}\right)$.
(I) Prove that the sequence $\left\{b_{n}\right\}$ is a geometric sequence, and find the general term formula for the sequence $\left\{b_{n}\right\}$;
(II) Let $c_{n}=-n b_{n}\left(n \in N^{*}\right)$, find the sum of the first $n$ terms of the sequence $\left\{c_{n}\right\}$, denoted as $T_{n}$.
Solution: (1) $\frac{x_{n+1}-1}{x_{n+1}+1}=\frac{f\left(x_{n}\right)-1}{f\left(x_{n}\right)+1}=\frac{\frac{x_{n}^{3}+3 x_{n}}{3 x_{n}^{2}+1}-1}{\frac{x_{n}^{3}+3 x_{n}}{3 x_{n}^{2}+1}+1}=\frac{x_{n}^{3}-3 x_{n}^{2}+3 x_{n}-1}{x_{n}^{3}+3 x_{n}^{2}+3 x_{n}+1}=\left(\frac{x_{n}-1}{x_{n}+1}\right)^{3}$
(5 points)
Thus, $\log _{3}\left(\frac{x_{n+1}-1}{x_{n+1}+1}\right)=3 \log _{3}\left(\frac{x_{n}-1}{x_{n}+1}\right)$, i.e., $b_{n+1}=3 b_{n}$, so the sequence $\left\{b_{n}\right\}$ is a geometric sequence. Also, $b_{1}=\log _{3}\left(\frac{2-1}{2+1}\right)=-1$, thus $b_{n}=-3^{n-1}$. Therefore, the general term formula of the sequence $\left\{b_{n}\right\}$ is $b_{n}=-3^{n-1}$.
(10 points)
(II) From (I), we know that $b_{n}=-3^{n-1}$, hence $c_{n}=n \cdot 3^{n-1}$,
$$
\begin{array}{c}
T_{n}=1 \times 3^{0}+2 \times 3^{1}+3 \times 3^{2}+\cdots+n \times 3^{n-1}, \\
3 T_{n}=\quad 1 \times 3^{1}+2 \times 3^{2}+3 \times 3^{3}+\cdots+n \times 3^{n},
\end{array}
$$

Thus, $-2 T_{n}=1+3+3^{2}+\cdots+3^{n-1}-n \times 3^{n}=\frac{3^{n}-1}{2}-n \times 3^{n}$,
i.e., $T_{n}=\frac{n \times 3^{n}}{2}-\frac{3^{n}-1}{4}=\frac{(2 n-1) \cdot 3^{n}+1}{4}$,

Therefore, the sum of the first $n$ terms of the sequence $\left\{c_{n}\right\}$ is $T_{n}=\frac{(2 n-1) \cdot 3^{n}+1}{4}\left(n \in N^{*}\right)$.