To identify the interval where \( f(x) = \frac{3}{x} + \frac{x}{3} \) is strictly decreasing, we must first compute its derivative, \( f'(x) \), and then analyze the sign of \( f'(x) \).
The derivative of \( f(x) \) is calculated as \( f'(x) = -\frac{3}{x^2} + \frac{1}{3} \).
The function is strictly decreasing when \( f'(x)<0 \). Therefore, we set up the inequality:
\(-\frac{3}{x^2} + \frac{1}{3}<0\).
To eliminate the fractions, we multiply the inequality by \( x^2 \):
\(-3 + \frac{x^2}{3}<0\).
Further simplification by multiplying by \( 3 \) yields:
\(-9 + x^2<0\).
Rearranging this inequality gives:
\( x^2<9 \).
Solving for \( x \), we find:
\(-3<x<3\).
Considering the domain of \( f(x) \), which excludes \( x = 0 \), the interval where \( f(x) \) is strictly decreasing is the union of two open intervals:
\((-3, 0) \cup (0, 3)\).