14. Given the function $f(x)=\frac{x^{3}+3 x}{3 x^{2}+1}$, the sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=2, x_{n+1}=f\left(x_{n}\right)\left(n \in \mathbf{N}_{+}\right)$.
Let $b_{n}=\log _{3}\left(\frac{x_{n+1}-1}{x_{n+1}+1}\right)\left(n \in \mathbf{N}_{+}\right)$.
(1) Prove that $\left\{b_{n}\right\}$ is a geometric sequence, and find the general term formula for the sequence $\left\{b_{n}\right\}$;
(2) Let $c_{n}=-n b_{n}\left(n \in \mathbf{N}_{+}\right)$, find the sum of the first $n$ terms of the sequence $\left\{c_{n}\right\}$, denoted as $T_{n}$.
14. (1) From the problem, we have
$$
\begin{array}{l}
\frac{x_{n+1}-1}{x_{n+1}+1}=\frac{f\left(x_{n}\right)-1}{f\left(x_{n}\right)+1} \\
=\frac{x_{n}^{3}-3 x_{n}^{2}+3 x_{n}-1}{x_{n}^{3}+3 x_{n}^{2}+3 x_{n}+1}=\left(\frac{x_{n}-1}{x_{n}+1}\right)^{3} .
\end{array}
$$

Thus, $\log _{3} \frac{x_{n+1}-1}{x_{n+1}+1}=3 \log _{3} \frac{x_{n}-1}{x_{n}+1}$, which means
$$
b_{n}=3 b_{n-1} \text {. }
$$

Therefore, $\left\{b_{n}\right\}$ is a geometric sequence.
Given $x_{2}=\frac{14}{13}$, then
$$
b_{1}=\log _{3} \frac{x_{2}-1}{x_{2}+1}=\log _{3} \frac{1}{27}=-3 .
$$

Thus, $b_{n}=-3^{n}$.
(2) From (1.), we know $b_{n}=-3^{n}$. Therefore, $c_{n}=3^{n} n$.

Then $T_{n}=3^{1} \times 1+3^{2} \times 2+\cdots+3^{n} n$
$$
\begin{aligned}
\Rightarrow & 3 T_{n}=3^{2} \times 1+3^{3} \times 2+\cdots+3^{n+1} n \\
\Rightarrow & -2 T_{n}=3+3^{2}+\cdots+3^{n}-3^{n+1} n \\
& =\frac{3^{n+1}-3}{2}-3^{n+1} n
\end{aligned}
$$
$$
\Rightarrow T_{n}=\frac{3^{n+1}(2 n-1)+3}{4} .
$$