It is ok to use the maclaurin series for $\frac{1}{x+1} $ to expand the function $$f(x) = \frac{x}{x^4+x^2+1}$$ like that: $$f(x) = x\frac{1}{(x^4+x^2)+1} = x\sum_{n=0}(-1)^n(x^4+x^2)^n = \sum_{n=0}(-1)^nx(x^4+x^2)^n $$
$$=\sum_{n=0}(-1)^nx^{2n+1} (x^2+1)^n$$
Hint:
For $x^2-1\ne0,$
$$\dfrac x{1+x^2+x^4}=x(1-x^2)\cdot(1-x^6)^{-1}$$