Solution:

Combining fractions, $f_1(x) = \frac{2(3x + 1) - 9}{3(3x + 1)} = \frac{6x - 7}{9x + 3}.$ Composing once, $f_2(x) = \frac{6 f_1(x) - 7}{9 f_1(x) + 3} = \frac{-3x - 7}{9x - 6},$ and composing again gives $f_3(x) = x.$

So the iteration is periodic with period $3.$ Since $1001 \equiv 2 \pmod{3},$ we have $f_{1001} = f_2,$ and the equation becomes $$\frac{-3x - 7}{9x - 6} = x - 3,$$ that is $9x^2 - 33x + 18 = -3x - 7,$ or $9x^2 - 30x + 25 = (3x - 5)^2 = 0.$

The unique solution is $x = \frac{5}{3},$ so $m + n = 5 + 3 = 8.$