$f(x)=?$ if we have $f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$ $f(x)=?$
If we have $$f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}$$ to fractions are very similar. I don't have an idea to find $f(x)$. Can someone show me a clue ?
 A: Hint:
$$f\left(\frac{x}{x^2+x+1}\right)=\frac{x}{x^2-x+1}=\frac{1}{\frac{x^2-x+1}{x}}=\frac{1}{\frac{x^2+x+1-2x}{x}}=\frac{1}{\frac{x^2+x+1}{x}-2}$$
Note, that we have to assume $x\neq 0$ in this process. You will have to check if that is problematic.
A: HInt: you can take $$x+\frac 1x  =u $$ and turn all the expression into $u$ ,or do like below 
$$\quad{f(\frac{x}{x^2+x+1})=\frac{x}{x^2-x+1}\\
\frac{x}{x^2+x+1}=u\\\frac{x^2+x+1}{x}=\frac 1u\\x+\frac{1}x +1=\frac 1u \\\to x+\frac{1}x =\frac 1u -1\\
\frac{x}{x^2-x+1}=\dfrac{1}{\dfrac{x^2-x+1}{x}}=\\\dfrac{1}{x+\dfrac{1}{x}-1}=\dfrac{1}{(\frac 1u -1)-1}}$$can you go on  ?
A: Let $y = \frac x{x^2+x+1}$ and $z = \frac x{x^2-x+1}$. Then we have
\begin{align}
y(x^2+x+1) = x &\implies y(x^2-x+1)=x -2xy\\
&\implies y=z(1-2y)\\
&\implies z=\frac y{1-2y}\\
\end{align}
which gives us $f(y) = \frac{y}{1-2y}$.
A: $$f\left( \dfrac{x}{x^{2}+x+1}\right)=\dfrac{x}{x^{2}-x+1}\rightarrow f\left( \dfrac{1}{x+\frac{1}{x}+1}\right)=\dfrac{1}{x+\frac{1}{x}-1}\rightarrow$$
$$f\left( \dfrac{1}{z+1}\right)=\dfrac{1}{z-1}$$
$$\dfrac{1}{z+1}=t\rightarrow \dfrac{1}{t}=z+1\rightarrow z=\dfrac{1}{t}-1\rightarrow z=\dfrac{1-t}{t}\rightarrow$$
$$f(t)=\dfrac{1}{\frac{1-t}{t}-1}=\dfrac{1}{\frac{1-2t}{t}}=\dfrac{t}{1-2t}\rightarrow f(x)=\dfrac{x}{1-2x}$$
Testing:
$$f\left( \frac{x}{x^{2}+x+1}\right)=\dfrac{\frac{x}{x^{2}+x+1}}{1-2\left( \frac{x}{x^{2}+x+1}\right)}=\dfrac{\frac{x}{x^{2}+x+1}}{\frac{x^{2}+x+1-2x}{x^{2}+x+1}}=\dfrac{x}{x^{2}-x+1}$$
