Let $h(x) = \frac{x}{x^2 + x + 1}$. Taking the first derivative equal to zero yields the critical points: $h'(x) = \frac{(x^2 + x + 1) - x(2x + 1)}{(x^2 + x +1)^2} = \frac{-x^2 + 1}{(x^2 + x + 1)^2} = 0 \Rightarrow x = \pm 1$ and the second derivative of $h(x)$ at these critical points yields: $h''(x) = \frac{(-2x)(x^2 + x + 1)^2 - (1 - x^2)*2(x^2 + x + 1)(2x + 1)}{(x^2 + x + 1)^4} = \frac{2(x^3 - 3x - 1)}{(x^2 + x + 1)^3}$ or $h''(1) = -2/9 < 0$ (a maximum) and $h''(-1) = 2 > 0$ (a minimum). Hence $h(1) = 1/3$ and $h(-1) = -1$. Next, we wish to determine all functions $f(x)$ such that $f(x/(x^2 + x +1)) = \frac{x^2}{x^4 + x^2 + 1}$ for all real $x$. Upon observation, we find that: $\frac{x^2}{x^4 + x^2 + 1} = \frac{x/(x^2 + x + 1)}{x/(x^2 - x + 1)}$; or $\frac{x/(x^2 + x + 1)}{(x^2 - x + 1)/x}$; or $\frac{x/(x^2 + x + 1)}{x^2 + x + 1 - 2x)/x}$; or $\frac{x/(x^2 + x + 1)}{(x^2 + x + 1)/x - 2}$. If $y = \frac{x}{x^2 + x + 1}$, then $f(y) = \frac{y}{1/y - 2} \Rightarrow f(y) = \frac{y^2}{1 - 2y}$.

(a) \[ -1 \leq \frac{x}{x^2 + x + 1} \implies -(x^2 + x + 1)^2 \leq x(x^2 + x + 1) \implies (x^2 + x + 1)(x^2 + 2x + 1) \geq 0\]which is true \[ \frac{x}{x^2 + x + 1} \leq \frac{1}{3} \implies x(x^2 + x + 1) \leq \frac{1}{3} (x^2 + x + 1)^2 \implies (x^2 + x + 1)(x^2 - 2x + 1) \geq 0 \]which is true (b)\[\frac{x^2}{x^4 + x^2 + 1}=\frac{x}{x^2 + x + 1} \frac{x}{x^2 - x + 1} = y \cdot \frac{1}{\frac{1}{y}-2} = \frac{y^2}{1-2y}\]Thus, $f(x)=\frac{x^2}{1-2x}$