<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] $f(f(x)) =f^{-1}(1)$
$$f(x) = x-\frac{1}{x}$$
$$f^{-1}(x) = \frac{x+\sqrt{x^2+4}}{2}$$
or
$$f^{-1}(x) = \frac{x-\sqrt{x^2+4}}{2}$$
It means we have 2 possible values of $f^{-1}(1)$ 
$f^{-1}(1)=\phi_1=\frac{1+\sqrt{5}}{2}$ or $f^{-1}(1)=\phi_2=\frac{1-\sqrt{5}}{2}$
For $\phi_1=\frac{1+\sqrt{5}}{2}$
$f(f(x)) =\phi_1=\frac{1+\sqrt{5}}{2}$
As you pointed in question we know that $f(f(x)) = x-\frac{1}{x}-\frac{1}{x-\frac{1}{x}} $
$x-\frac{1}{x}=P$  then
$f(f(x)) = P-\cfrac{1}{P}= \phi_1$
You can solve $P$
$P^2-\phi_1P-1=0$
You will find 2 values from here and you will use in it
$x-\frac{1}{x}=P_1$  and $x-\frac{1}{x}=P_2 $ then you will find 4 different x values.
Then you will do the same things for  For $\phi_2=\frac{1-\sqrt{5}}{2}$,
$f(f(x)) = P-\cfrac{1}{P}= \phi_2$ and you will find other 4 different x values
Totally 8 different x , you will find.
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