$\frac{x+5}{x^{2}-9}+\frac{x-5}{2 x-6}=\frac{x}{x+3}$

Step 1 ：\(\frac{x+5}{x^{2}-9} + \frac{x-5}{2x-6} = \frac{x}{x+3}\)

Step 2 ：Combine the fractions: \(\frac{(x+5)(2x-6)}{(x^{2}-9)(2x-6)} + \frac{(x-5)(x^{2}-9)}{(2x-6)(x^{2}-9)} = \frac{x(x^{2}-9)(2x-6)}{(x+3)(x^{2}-9)(2x-6)}\)

Step 3 ：Hence, \((x+5)(2x-6)(x+3) + (x-5)(x^{2}-9)(x+3) = x(x^{2}-9)(2x-6)\)

Step 4 ：Simplify: \(3x(x+3)(x+5) = x(x^{2}-9)(2x-6)\)

Step 5 ：Divide both sides by x(x+3): \(3(x+5) = (x^{2}-9)(2x-6)\)

Step 6 ：Expand: \(3x+15 = 2x^{3}-18x^{2}+12x^{2}-108x+18x-54\)

Step 7 ：Simplify: \(2x^{3}-6x^{2}-123x-39 = 0\)

Step 8 ：Factor out a common factor of 3: \(2x^{3}-6x^{2}-123x-39 = 3(2x^{3}-2x^{2}-41x-13) = 0\)

Step 9 ：Thus, \(2x^{3}-2x^{2}-41x-13 = 0\)

Step 10 ：The final answer is \(\boxed{2x^{3}-2x^{2}-41x-13}\)