\(\frac{1}{3+\sqrt{2}}+\frac{1}{3-\sqrt{2}}=\frac{3-\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}+\frac{3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\frac{3-\sqrt{2}+3+\sqrt{2}}{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}=\frac{6}{3^2-2}=\frac{6}{7}\Rightarrowđpcm\)