已知$f(x)$是定义在$(-1,1)$上的函数，$f\left(\dfrac 12\right)=-1$，且对任意$x,y\in (-1,1)$，有$f(x)+f(y)=f\left(\dfrac{x+y}{1+xy}\right)$．
(1) 求证：$f(x)$是奇函数；
(2) 若数列$\{x_n\}$满足$x_1=\dfrac 12$，$x_{n+1}=\dfrac{2x_n}{1+x_n^2}$，求$f(x_n)$；
(3) 证明：$1+f\left(\dfrac 15\right)+f\left(\dfrac 1{11}\right)+\cdots +f\left(\dfrac{1}{n^2+3n+1}\right)+f\left(\dfrac{1}{n+2}\right)=0$．
分析与解 (1) 令$x=y=0$，可得$2f(0)=f(0)$，于是$f(0)=0$．再令$y=-x$，则有$$f(x)+f(-x)=f(0)=0,$$于是命题得证．
(2) 令$x=y=x_n$可得$$2f(x_n)=f\left(\dfrac{2x_n}{1+x_n^2}\right)=f(x_{n+1}),$$于是$\{f(x_n)\}$是首项为$f\left(\dfrac 12\right)=-1$，公比为$2$的等比数列，进而$$f(x_n)=-2^{n-1}(n\in\mathbb N^*).$$(3) 根据题意，有\[\begin{split} f\left(\dfrac{1}{k^2+3k+1}\right)&=f\left(\dfrac{(k+2)-(k+1)}{(k+2)\cdot (k+1)-1}\right)\\&=f\left(\dfrac{\dfrac{1}{k+1}-\dfrac{1}{k+2}}{1-\dfrac{1}{k+1}\cdot \dfrac{1}{k+2}}\right)\\&=f\left(\dfrac{1}{k+1}\right)+f\left(-\dfrac{1}{k+2}\right)\\&=f\left(\dfrac{1}{k+1}\right)-f\left(\dfrac{1}{k+2}\right),\end{split} \]于是$$\sum_{k=1}^{n}f\left(\dfrac{1}{k^2+3k+1}\right)=\sum_{k=1}^{n}\left[f\left(\dfrac{1}{k+1}\right)-f\left(\dfrac{1}{k+2}\right)\right]=f\left(\dfrac 12\right)-f\left(\dfrac{1}{n+2}\right),$$因此原命题得证．