To solve the equation

\[ \frac{3}{3x+1} + \frac{2}{1-3x} = \frac{x-5}{9x^2-1}, \]

we start by simplifying the right-hand side of the equation.

Notice that \( 9x^2 - 1 \) can be factored as a difference of squares:

\[ 9x^2 - 1 = (3x + 1)(3x - 1). \]

So the equation becomes:

\[ \frac{3}{3x+1} + \frac{2}{1-3x} = \frac{x-5}{(3x+1)(3x-1)}. \]

Next, we need to find a common denominator for the fractions on the left-hand side. Notice that:

\[ 1 - 3x = -(3x - 1). \]

So we rewrite the second fraction:

\[ \frac{2}{1-3x} = \frac{2}{-(3x-1)} = -\frac{2}{3x-1}. \]

Now the equation is:

\[ \frac{3}{3x+1} - \frac{2}{3x-1} = \frac{x-5}{(3x+1)(3x-1)}. \]

The common denominator for the left-hand side is \((3x+1)(3x-1)\). Rewrite the fractions with this common denominator:

\[ \frac{3(3x-1)}{(3x+1)(3x-1)} - \frac{2(3x+1)}{(3x+1)(3x-1)} = \frac{x-5}{(3x+1)(3x-1)}. \]

Simplify the numerators:

\[ \frac{3(3x-1) - 2(3x+1)}{(3x+1)(3x-1)} = \frac{x-5}{(3x+1)(3x-1)}. \]

Expand and combine like terms in the numerator:

\[ \frac{9x - 3 - 6x - 2}{(3x+1)(3x-1)} = \frac{x-5}{(3x+1)(3x-1)}. \]

\[ \frac{3x - 5}{(3x+1)(3x-1)} = \frac{x-5}{(3x+1)(3x-1)}. \]

Since the denominators are the same, we can equate the numerators:

\[ 3x - 5 = x - 5. \]

Subtract \(x\) from both sides:

\[ 2x - 5 = -5. \]

Add 5 to both sides:

\[ 2x = 0. \]

Divide by 2:

\[ x = 0. \]

So the solution to the equation is:

\[ x = 0. \]

We should check if \( x = 0 \) is valid by substituting it back into the original equation to ensure it does not make any denominator zero. Substituting \( x = 0 \):

\[ \frac{3}{3(0)+1} + \frac{2}{1-3(0)} = \frac{0-5}{9(0)^2-1}, \]

\[ \frac{3}{1} + \frac{2}{1} = \frac{-5}{-1}, \]

\[ 3 + 2 = 5, \]

\[ 5 = 5. \]

So \( x = 0 \) is indeed a valid solution.