<s>[INST] if $f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$ If $\displaystyle f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$
$\underline{\bf{My\;\; Try}}::$ Given $\displaystyle f(x) = x-\frac{1}{x} = \frac{x^2-1}{x}.$ Now Replace $\displaystyle x\rightarrow \frac{1}{x}\;,$ We Get.
$\displaystyle f(f(x)) = x-\frac{1}{x}-\frac{x}{x^2-1} = \frac{x^2-1}{x}-\frac{x}{x^2-1} = \frac{(x^2-1)^2-x^2}{x.(x^2-1)} = \frac{x^4-3x^2+1}{x.(x^2-1)}$
again Replace $\displaystyle x\rightarrow \frac{1}{x}\;,$ We Get
$\displaystyle f(f(f(x))) = \frac{\left(\frac{x^2-1}{x}\right)^4-3\left(\frac{x^2-1}{x}\right)^2+1}{\left(\frac{x^2-1}{x}\right).\left\{\left(\frac{x^2-1}{x}\right)^2-1\right\}}$
Now I did not understand how can I solve This $8^{th}$ Degree equation,
So Help Required,
Thanks
 [/INST] You want to solve $f(f(f(x)))=1$ for $x$. You can solve it by solving three equations in turn: Set $a:=f(f(x))$, then find $a$ via $f(a)=1$. Then set $b:=f(x)$ and find $b$ via $f(b)=a$. Lastly, find $x$ via $f(x)=b$. Notice also that because $f(-\frac{1}{x_0})=f(x_0)$, for each solution $x_0$, the number $-\frac{1}{x_0}$ is a solution too.
There are $8$ solutions and they all lie in the interval $(-1.9,2.5)$. You can e.g. take Wolfram alpha and plot your ugly rational function and see how often it takes the value $1$. Solving the equation is a one-liner in a program like Mathematica. One solution is $x=-0.40065\dots$, convince yourself by plugging in $-0.4$ and you find $f(f(f(-0.4)))=1.00797\dots$. 
</s>