\(\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}=\dfrac{3-\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+\dfrac{3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{\left(3-\sqrt{5}\right)^2}=\dfrac{6}{9-5}=\dfrac{6}{4}=\dfrac{3}{2}\)