To solve the given problem, we need to analyze each statement individually within the given range for \(x\) and \(y\):

The problem constraints are \(\frac{1}{3} < x < y < 2\).

Statement I:

The statement is \(x + \frac{1}{x} < y + \frac{1}{y}\).

Given \(x < y\), let's evaluate if this condition is always true:

• We define a function \(f(a) = a + \frac{1}{a}\).
• The derivative of this function \(f'(a) = 1 - \frac{1}{a^2}\).
• For \(a > 1\), \(f'(a) > 0\) which means \(f(a)\) is increasing.
• For \(\frac{1}{3} < a < 1\), observe that it falls within the interval where \(f(a)\) increases. Therefore, \(x + \frac{1}{x} < y + \frac{1}{y}\) holds true because \(f(a)\) is increasing with \(a > 1\) and within our range.

Thus, Statement I is always correct.

Statement II:

The statement is \(\frac{\sqrt{1+y^2}}{y} < \frac{\sqrt{1+x^2}}{x}\).

We need to check if this inequality holds for \(\frac{1}{3} < x < y < 2\):

• Consider the function \(g(a) = \frac{\sqrt{1+a^2}}{a}\).
• The derivative is complex, but note that as \(a\) increases, \(\frac{\sqrt{1+a^2}}{a}\) decreases for \(a > 1\) due to the fact that the increase in \(a\) makes the numerator and denominator relatively stable.
• This implies that \(g(y) \lt g(x)\) would not generally hold for all \(x, y\) in this range.

Thus, Statement II is not correct for all values within the specified range.

Based on the above reasoning, the correct answer is I only.