$\frac{3+\mathrm{i}}{(3-\mathrm{i})^{2}}+\frac{3-\mathrm{i}}{(3+\mathrm{i})^{2}}=\frac{9}{25}$

Step 1 ：\(\frac{3+i}{(3-i)^2} + \frac{3-i}{(3+i)^2}\)

Step 2 ：Multiply the first term by \(\frac{3+i}{3+i}\) and the second term by \(\frac{3-i}{3-i}\)

Step 3 ：\(\frac{(3+i)^2}{(3-i)^2(3+i)} + \frac{(3-i)^2}{(3+i)^2(3-i)}\)

Step 4 ：Combine the terms with a common denominator

Step 5 ：\(\frac{(3+i)^2(3-i) + (3-i)^2(3+i)}{(3-i)^2(3+i)^2}\)

Step 6 ：Expand the numerator

Step 7 ：\(\frac{9+6i+i^2+9-6i-i^2}{(3-i)^2(3+i)^2}\)

Step 8 ：Simplify the numerator using \(i^2=-1\)

Step 9 ：\(\frac{18}{(3-i)^2(3+i)^2}\)

Step 10 ：Expand the denominator

Step 11 ：\(\frac{18}{(9-6i+1)(9+6i+1)}\)

Step 12 ：Simplify the denominator

Step 13 ：\(\frac{18}{100}\)

Step 14 ：Reduce the fraction

Step 15 ：\(\boxed{\frac{9}{50}}\)