1. **State the problem:** We want to prove the upper bound of the function $$f(x) = \frac{x}{x^2 - x}$$ for $$x \in (2, \infty)$$. 2. **Rewrite the function:** Simplify the denominator: $$ f(x) = \frac{x}{x^2 - x} = \frac{x}{x(x-1)} $$ 3. **Cancel common factors:** Since $$x > 2 > 0$$, we can cancel $$x$$ safely: $$ f(x) = \frac{\cancel{x}}{\cancel{x}(x-1)} = \frac{1}{x-1} $$ 4. **Analyze the simplified function:** The function reduces to $$f(x) = \frac{1}{x-1}$$ for $$x > 2$$. 5. **Determine the upper bound:** Since $$x > 2$$, the denominator $$x-1 > 1$$, so: $$ f(x) = \frac{1}{x-1} < 1 $$ 6. **Conclusion:** The function $$f(x)$$ is strictly less than 1 for all $$x \in (2, \infty)$$, so the upper bound is 1. **Final answer:** $$\boxed{f(x) < 1 \text{ for } x \in (2, \infty)}$$