The greatest value of  $f\left(x\right)=\frac{\{{(x+\frac{1}{x})}^4-(x^4+\frac{1}{x^4})-1\}}{{(x+\frac{1}{x})}^2+(x^2+\frac{1}{x^2})}$  for   $\forall x\in R-\left\{0\right\}$  is

$$\begin{gathered} \frac{{(x+\frac{1}{x})}^4-(x^4+\frac{1}{x^4})-1}{{(x+\frac{1}{x})}^2+(x^2+\frac{1}{x^2})} \\ Put x+\frac{1}{x}=1 \Rightarrow y=\frac{t^4-[{(t^2-2)}^2-2]-1}{t^2+t^2-2}=\frac{t^4-[t^4-4t^2+2]-1}{2(t^2-1)} \\ \Rightarrow y=\frac{4t^2-3}{2(t^2-1)}\Rightarrow y=\frac{4t^2-4+1}{2(t^2-1)}=(2+\frac{1}{2(t^2-1)}) \\ As t=x+\frac{1}{x}\Rightarrow t^2\ge 4\Rightarrow t^2-1\ge 3\Rightarrow \frac{1}{t^2-1}\le \frac{1}{3} \end{gathered}$$

$\Rightarrow$ Maximum value of y is $2+\frac{1}{3\times 2}=\frac{13}{6}$