Don't use calculus

$\begin{aligned} \frac{\left(x+\frac{1}{x}\right)^6 - \color{#3D99F6}{ \left(x^6+\frac{1}{x^6}\right)} -2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)} & = \frac{\left(x+\frac{1}{x}\right)^6 - \color{#3D99F6}{\left[ \left(x^3+\frac{1}{x^3}\right)^2 - 2\right]} -2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)} \\ & = \frac{\left(x+\frac{1}{x}\right)^6 - \left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)} \\ & = \frac{\left[ \left(x+\frac{1}{x}\right)^3 - \left(x^3+\frac{1}{x^3}\right)\right] \left[ \left(x+\frac{1}{x}\right)^3 + \left(x^3+\frac{1}{x^3}\right)\right]}{\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)} \\ & = \left(x+\frac{1}{x}\right)^3 - \left(x^3+\frac{1}{x^3}\right) \\ & = x^3+3x+\frac{3}{x}+\frac{1}{x^3} - x^3-\frac{1}{x^3} \\ & = 3\left(x+\frac{1}{x}\right) \end{aligned}$

Using AM-GM inequality for $x > 0$ , we have $x+\dfrac{1}{x} \ge 2$ .

Then the maximum $\dfrac{\left(x+\frac{1}{x}\right)^6 - \left(x^6+\frac{1}{x^6}\right) -2}{\left(x+\frac{1}{x} \right)^3 + \left( x^3+\frac{1}{x^3}\right)} = 3 \times 2 =\boxed{6}$

Let $\boxed{(x+\frac{1}{x})=m}.......(1)$ . Therefore $(x^{3}+\frac{1}{x^{3}})+3(x+\frac{1}{x})=m^{3}$ $\boxed{(x^{3}+\frac{1}{x^{3}})=m^{3}-3m}.......(2)$ Squaring both sides , we get $\boxed{(x^{6}+\frac{1}{x^{6}})=m^{6}+9m^{2}-6m^{4}-2}........(3)$ Now rewriting the given expression in terms of $m$ we get $=\frac{m^{6}-(m^{6}+9m^{2}-6m^{4}-2)-2}{m^{3}+m^{3}-3m}$ $=\frac{6m^{4}-9m^{2}}{2m^{3}-3m}$ $=3m$ By $A.M-G.M$ $x+\frac{1}{x}\geq2$ Therefore $3m\geq6$ .

Hence the minimum value of the given expression is $6$