1. **State the problem:** We want to prove the upper bound of the function $$f(x) = \frac{x}{x^2 - 1}$$ for $$x \in (2, \infty)$$. 2. **Analyze the function:** The function is defined for $$x > 1$$ since the denominator $$x^2 - 1$$ is zero at $$x=\pm 1$$ and undefined there. 3. **Find the derivative to locate extrema:** Use the quotient rule: $$f'(x) = \frac{(1)(x^2 - 1) - x(2x)}{(x^2 - 1)^2} = \frac{x^2 - 1 - 2x^2}{(x^2 - 1)^2} = \frac{-x^2 - 1}{(x^2 - 1)^2}$$ 4. **Simplify the numerator:** $$-x^2 - 1 = -(x^2 + 1)$$ which is always negative for all real $$x$$. 5. **Determine the sign of the derivative:** Since the denominator $$(x^2 - 1)^2 > 0$$ for all $$x \neq \pm 1$$ and numerator is negative, $$f'(x) < 0$$ for all $$x > 1$$. 6. **Interpretation:** The function $$f(x)$$ is strictly decreasing on the interval $$(2, \infty)$$. 7. **Find the upper bound:** Since $$f(x)$$ is decreasing on $$(2, \infty)$$, its maximum value on this interval is the limit as $$x$$ approaches 2 from the right: $$\lim_{x \to 2^+} \frac{x}{x^2 - 1} = \frac{2}{2^2 - 1} = \frac{2}{4 - 1} = \frac{2}{3}$$ 8. **Conclusion:** The function $$f(x) = \frac{x}{x^2 - 1}$$ for $$x \in (2, \infty)$$ is strictly decreasing and bounded above by $$\frac{2}{3}$$. **Final answer:** $$\boxed{f(x) \leq \frac{2}{3} \text{ for all } x \in (2, \infty)}$$