Hmm...alternatively you can let $\displaystyle x = \frac{\cot \theta}{\sqrt{3}}$, you should still obtain the same thing.
Daigo has just taken Algebra. Soon, i think the suggests should somewhat easier to understand; not calculus or trigonometry, but basic Algebra.
You should estimate the points on the graph that are max. and min.
Aha! I thought there would be an non-calculus, non-trig, algebraic solution to this problem.
Let k be the maximal value of $\displaystyle \frac{2x}{3x^2 + 1} $ so that $\displaystyle \frac{2x}{3x^2 + 1} = k$. Cross-multiplying, we obtain
$\displaystyle 3kx^2 - 2x + k = 0 \Rightarrow x = \frac{2 \pm \sqrt{4 - 4(3k)(k)}}{6k} = \frac{2 \pm \sqrt{4 - 12k^2}}{6k}$. We want k to be as large as possible subject to the constraint that x is real. This implies that
$\displaystyle 4 - 12k^2 \ge 0 \Rightarrow k^2 \le \frac{1}{3}$. The largest possible value of k is $\displaystyle \sqrt{\frac{1}{3}}$.
Furthermore, since $\displaystyle y = \frac{2x}{3x^2 + 1}$ is an odd function, the smallest possible value of k is $\displaystyle -\sqrt{\frac{1}{3}}$.