Find the greatest value of $f\left(x\right)=\frac{{\left(x+\frac{1}{x}\right)}^4-\left(x^4+\frac{1}{x^4}\right)-1}{{\left(x+\frac{1}{x}\right)}^2+\left(x^2+\frac{1}{x^2}\right)},x\in R-\left\{0\right\}$

Let 

$\begin{array}{l}g\left(x\right)={\left(x+\frac{1}{x}\right)}^4-\left(x^4+\frac{1}{x^4}\right)-1\\ ={\left\{{\left(x+\frac{1}{x}\right)}^2\right\}}^2-\left(x^4+\frac{1}{x^4}\right)-1\\ ={\left(x^2+\frac{1}{x^2}+2\right)}^2-\left(x^4+\frac{1}{x^4}\right)-1\\ ={\left(x^2+\frac{1}{x^2}\right)}^2+2\left(x^2+\frac{1}{x^2}\right)-\left(x^4+\frac{1}{x^4}\right)+3\\ =2\left(x^2+\frac{1}{x^2}\right)+5\end{array}$

and 

$\begin{array}{l}h\left(x\right)={\left(x+\frac{1}{x}\right)}^2+\left(x^2+\frac{1}{x^2}\right)\\ =2\left(x^2+\frac{1}{x^2}\right)+3\end{array}$

Thus, 

$\begin{array}{l}f\left(x\right)=\frac{g\left(x\right)}{h\left(x\right)}=\frac{2\left(x^2+\frac{1}{x^2}\right)+5}{2\left(x^2+\frac{1}{x^2}\right)+3}\\ =1+\frac{2}{2\left(x^2+\frac{1}{x^2}\right)+3}\\ =1+\frac{2}{2{\left(x+\frac{1}{x}\right)}^2-1}\end{array}$

It will be maximum when x = - 1

Hence the maximum value of f{x) is $\frac{9}{7}$