The interval, in which the function $f(x)=\frac{3}{x}+\frac{x}{3}$ is strictly decreasing, is:

$(-3,0)∪(0,3)$

The correct answer is Option (3) → $(-3,0)∪(0,3)$

$f(x)=\frac{3}{x}+\frac{x}{3}$

for $f(x)$ to be strictly decreasing, $f'(x)<0$

$⇒\frac{-3}{x^2}+\frac{1}{3}<0$

$⇒\frac{-9+x^2}{3x^2}<0$

$⇒-9+x^2<0$

$⇒x<+3$

x>-3

as $x≠0$ as $\frac{3}{x}$ is not defined.