$\text{ Let }f\left(x\right)=\frac{x-1}{x+1}\text{, }x\in R-\{0,-1,1)\text{. If }{f}^{n+1}\left(x\right)=f\left(f^2\left(x\right)\right)$ for all $n\in N$, then $f^6\left(6\right)+f^7\left(7\right)$ is equal to:

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}-1}{\mathrm{x}+1}\Rightarrow {\mathrm{f}}^2\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\frac{{\displaystyle \frac{x-1}{x+1}}-1}{\frac{\mathrm{x}-1}{\mathrm{x}+1}+1}=-\frac{1}{\mathrm{x}}\\ \\ {\mathrm{f}}^3\left(\mathrm{x}\right)=\mathrm{f}\left({\mathrm{f}}^2\left(\mathrm{x}\right)\right)=\mathrm{f}\left(-\frac{1}{\mathrm{x}}\right)=\frac{\mathrm{x}+1}{1-\mathrm{x}}\\ \Rightarrow {\mathrm{f}}^4\left(\mathrm{x}\right)=\mathrm{f}\left(\frac{\mathrm{x}+1}{1-\mathrm{x}}\right)=x\\ \Rightarrow {\mathrm{f}}^6\left(\mathrm{x}\right)=-\frac{1}{\mathrm{x}}\Rightarrow {\mathrm{f}}^6\left(6\right)=-\frac{1}{6}\\ {\mathrm{f}}^7\left(\mathrm{x}\right)=f\left(-\frac{1}{\mathrm{x}}\right)=\frac{\mathrm{x}+1}{1-\mathrm{x}}\\ \Rightarrow {\mathrm{f}}^7\left(7\right)=\frac{8}{-6}=-\frac{4}{3}\\ \therefore f^6\left(6\right)+f^7\left(7\right)=-\frac{1}{6}+-\frac{4}{3}=-\frac{3}{2}\end{array}$