\[\left(\frac{a}{3}+\frac{3}{a}\right)\frac{1}{a+3}=\frac{a^2+9}{3a(a+3)}.\]
При \(a=6\): \(\frac{36+9}{3\cdot6\cdot9}=\frac{45}{162}=\frac{5}{18}\).
Ответ: \(\frac{5}{18}\).