Let $f\left(x\right)=x^2+\frac{1}{x^2}\text{ and }g\left(x\right)=x-\frac{1}{x},x\in R-\{-1,0,1\}\text{. If }h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}\text{,}$ then the local minimum value of  $h\left(x\right)$ is

$\begin{array}{c}f\left(x\right)=x^2+\frac{1}{x^2}\text{ and }g\left(x\right)=x-\frac{1}{x}\Rightarrow h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}\\ \therefore h\left(x\right)=\frac{x^2+\frac{1}{x^2}}{x-\frac{1}{x}}=\frac{{(x-\frac{1}{x})}^2+2}{x-\frac{1}{x}}\\ \Rightarrow h\left(x\right)=(x-\frac{1}{x})+\frac{2}{x-\frac{1}{x}}\\ x-\frac{1}{x}>0,(x-\frac{1}{x})+\frac{2}{x-\frac{1}{x}}\in [2\sqrt{2},\infty )\\ x-\frac{1}{x}<0,(x-\frac{1}{x})+\frac{2}{x-\frac{1}{x}}\in (-\infty ,-2\sqrt{2}]\end{array}$

$\therefore \text{ Local minimum value is }2\sqrt{2}\text{. }$