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

We need to find $$f^6(6) + f^7(7)$$ where $$f(x) = \frac{x-1}{x+1}$$.

Since $$f^1(x) = \frac{x-1}{x+1}$$, a calculation shows $$f^2(x) = f(f(x)) = \frac{\frac{x-1}{x+1} - 1}{\frac{x-1}{x+1} + 1} = \frac{x - 1 - x - 1}{x - 1 + x + 1} = \frac{-2}{2x} = -\frac{1}{x}$$.

Then applying the function again gives $$f^3(x) = f\left(-\frac{1}{x}\right) = \frac{-\frac{1}{x} - 1}{-\frac{1}{x} + 1} = \frac{-1 - x}{-1 + x} = -\frac{x + 1}{x - 1}$$, and one more iteration yields $$f^4(x) = f\left(-\frac{x+1}{x-1}\right) = \frac{-\frac{x+1}{x-1} - 1}{-\frac{x+1}{x-1} + 1} = \frac{-x-1-x+1}{-x-1+x-1} = \frac{-2x}{-2} = x$$.

Therefore, the function has period 4: $$f^4(x) = x$$.

Now reducing the exponents modulo 4 gives $$f^6 = f^{4+2} = f^2$$ and $$f^7 = f^{4+3} = f^3$$.

Substituting these into the arguments yields $$f^6(6) = f^2(6) = -\frac{1}{6}$$ and $$f^7(7) = f^3(7) = -\frac{7+1}{7-1} = -\frac{8}{6} = -\frac{4}{3}$$.

From this it follows that $$f^6(6) + f^7(7) = -\frac{1}{6} - \frac{4}{3} = -\frac{1}{6} - \frac{8}{6} = -\frac{9}{6} = -\frac{3}{2}$$.

The correct answer is Option B.