Detailed Explanation:

Statement I: Define $f(t) = t + \frac{1}{t}$. Its derivative $f'(t) = 1 - \frac{1}{t^2}$ is negative on $(0,1)$ and positive on $(1,\infty)$. Thus $f(t)$ decreases on $(\frac{1}{3}, 1)$ and increases on $(1, 2)$—not monotonic throughout.

Counterexample: Take $x = \frac{1}{2}, y = \frac{3}{4}$. Then $x + \frac{1}{x} = 2.5$ but $y + \frac{1}{y} \approx 2.08$. So $x + \frac{1}{x} > y + \frac{1}{y}$, contradicting Statement I.

Statement II: Rewrite $\frac{\sqrt{1+t^2}}{t} = \sqrt{\frac{1}{t^2} + 1}$. For $t > 0$, as $t$ increases, $\frac{1}{t^2}$ decreases, making $\sqrt{\frac{1}{t^2}+1}$ strictly decreasing. Since $x < y$, we have $\frac{\sqrt{1+y^2}}{y} < \frac{\sqrt{1+x^2}}{x}$, confirming Statement II is always true.

Therefore, only Statement II is correct, making Option 2 the correct answer.