if $2f\left(\frac{x}{x^2+x+1}\right)=\frac{x^2}{x^4+x^2+1}$ then what is $f(x)$?

Question

assume that:

$$2f\left(\frac{x}{x^2+x+1}\right)=\frac{x^2}{x^4+x^2+1}$$

Then what is $f(x)$?

Answer 1

Assume that $u=\frac{x}{x^2+x+1}=\frac{1}{1+x+\frac{1}{x}}$. We have $\frac{1}{u}-1=x+\frac{1}{x}$, and by squaring: $$ \frac{1}{u^2}-\frac{2}{u} = x^2+\frac{1}{x^2}+1,$$ hence: $$ \frac{u^2}{1-2u} = \frac{1}{1+x^2+\frac{1}{x^2}} = \frac{x^2}{1+x^2+x^4} $$ and: $$ f(u) = \color{red}{\frac{u^2}{2-4u}}.$$

Answer 2

Since $$(x^2+x+1)^2=x^4+x^2+1+2x(x^2+x+1)$$ we have $$x^4+x^2+1=(x^2+x+1)^2-2x(x^2+x+1).$$ So,

$$\frac{x^2}{2(x^4+x^2+1)}=\frac{x^2}{2((x^2+x+1)^2-2x(x^2+x+1))}=\dfrac{\left(\dfrac{x}{x^2+x+1}\right)^2}{2-4\cdot\dfrac{x}{x^2+x+1}}$$