{(2x-1)^4 * 3(x+3)^2 - 4(2x-1)^3 * 2(x+3)^3} / (2x-1)^8

Easier if you start this way:

a = 2x - 1, b = x + 3 ; then:

(a^4 * 3b^2 - 4a^3 * 2b^3) / a^8

a^3(a * 3b^2 - 4 * 2b^3) / a^8 ... get my drift?

Then:

(a * 3b^2 - 4 * 2b^3) / a^5

(3ab^2 - 8b^3) / a^5

b^2(3a - 8b) / a^5

Now substitute back in.....

By the way, I see no reason why your teacher wouldn't be impressed if you submitted:

b^2(3a - 8b) / a^5 where a = 2x -1 and b = x + 3

Depends on the teacher, I guess....