Maximum value of function $f(x)=\frac{x^4-x^2}{x^6+2x^3-1}$ when $x >1$
Let $x-\frac{1}{x}=t$.
Thus, $t>0$ and by AM-GM we obtain: $$\frac{x^4-x^2}{x^6+2x^3-1}=\frac{x-\frac{1}{x}}{x^3-\frac{1}{x^3}+2}=\frac{t}{t^3+3t+2}=$$ $$=\frac{1}{t^2+\frac{2}{t}+3}\leq\frac{1}{3\sqrt[3]{t^2\left(\frac{1}{t}\right)^2}+3}=\frac{1}{6}.$$ The equality occurs for $t=1$, which says that we got a maximal value.