Number of real solutions of $f(f(f(x)))=1$, where $f(x) = x-x^{-1}$ given $f(x) = x-x^{-1}$ ,then number of real solution of $f(f(f(x)))=1$ 
$f(x)=\frac{x^2-1}{x}$, $f(f(x))=\frac{(f(x))^2-1}{f(x)} = \frac{(x^2-1)^2-x^2}{x(x^2-1)}=\frac{x^4-3x^2+1}{x^3-x}$
$f(f(f(x))) = \frac{(f(x))^4-3(f(x))^2+1}{(f(x))^3-f(x)} = \frac{(x^2-1)^4-3x^2(x^2-1)^2+x^4}{x[(x^2-1)^3-x^2(x^2-1)]}=1$
how can i solve 8 degree equation

Notice that $f$ is an odd function $f(-x)=-f(x)$.
We also have the relation $f(\frac 1x)=-f(x)=f(-x)$
From these two we can deduce $$f(f(f(-\frac 1x)))=f(f(f(x)))$$
Whenever $x$ is solution of $f^{\circ[3]}(x)=1$ then $-\frac 1x$ is solution too, so we can restrict our search to only positive roots.
The equation $f(x)=a$ is just a quadratic with $\Delta=a^2+4>0$ so the two roots of $f(x)=a$ are real and are given by $h(a)$ and $-\dfrac 1{h(a)}$.
Where $h(a)=\dfrac{a+\sqrt{a^2+4}}2\quad$ gives the positive root of the equation.
We can notice that $h(1)=\phi$ and $-\dfrac{1}{h(1)}=-\frac 1{\phi}=1-\phi$
Similarly we have $-\dfrac 1{h(a)}=a-h(a)$, this will be useful to express the roots of $g$ in a rationalized form (i.e. without square roots on the denominator).

We have the equation $$f(f(f(x)))=1\iff \begin{cases} f(z)=1\\f(y)=z\\f(x)=y\end{cases}$$
The $8$ solutions are $\{a,b,c,d,-\frac 1a,-\frac 1b,-\frac 1c,-\frac 1d\}$
$\begin{cases}
a=h(h(h(1)))\\
b=h(h(-1/h(1)))\\
c=h(-1/h(h(1)))\\
d=h(-1/h(-1/h(1)))
\end{cases}\quad$ or in the "rationalized" form $\quad\begin{cases}
a=h(h(h(1)))\\
b=h(h(1-h(1)))\\
c=h(h(1)-h(h(1)))\\
d=h(1-h(1)-h(1-h(1)))
\end{cases}$
This is more than enough to get numerical values for the roots:
$$a \approx 2.495944000\quad b\approx 1.434666438\quad c\approx 0.789447954\quad d\approx 0.530245300$$
Using the rationalized form you can get these formal solutions while plugging them into a CAS:
$\begin{cases}
a = \frac 18\left(1+\sqrt{5}+\sqrt{22+2\sqrt{5}}+\sqrt{92+4\sqrt{5}+2\sqrt{22+2\sqrt{5}}+2\sqrt{5}\sqrt{22+2\sqrt{5}}}\right)\\
b = \frac 18\left(1-\sqrt{5}+\sqrt{22-2\sqrt{5}}+\sqrt{92-4\sqrt{5}+2\sqrt{22-2\sqrt{5}}-2\sqrt{5}\sqrt{22-2\sqrt{5}}}\right)\\
c = \frac 18\left(1+\sqrt{5}-\sqrt{22+2\sqrt{5}}+\sqrt{92+4\sqrt{5}-2\sqrt{22+2\sqrt{5}}-2\sqrt{5}\sqrt{22+2\sqrt{5}}}\right)\\
d = \frac 18\left(1-\sqrt{5}-\sqrt{22-2\sqrt{5}}+\sqrt{92-4\sqrt{5}-2\sqrt{22-2\sqrt{5}}+2\sqrt{5}\sqrt{22-2\sqrt{5}}}\right)\\
\end{cases}$
