@`JEET
here, $f(x) = \frac{x-1}{x+1}$
Now, $f^2(x) = f(f(x)) = \frac{f(x)-1}{f(x)+1} = \frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1} + 1} = -1/x$
$f^3(x) =f(f^2(x) = f(-1/x) = \frac{-x-1}{x-1}$
$f^4(x) = f(f^3(x) = \frac{\frac{-x-1}{x-1}-1}{\frac{-x-1}{x-1} + 1} = x $
So, $f^4(x) = x$
It means $f^5(x) = f(x)$ and $f^6(x) = f^2(x)$ and $f^7(x) = f^3(x)$ and $f^8(x) = f^4(x) = x$
Now, observe the pattern, $f^4(x) = f^8(x) = f^{12}(x) = ......= f^{4k}(x) = x$ where $k\in I^+$
So, $f^{100}(x) = x$ which implies $f^{100}(10) = 10$