1. ## Chain Rule

We are supposed to differentiate the given function and simplify the answer. I'm not sure if I did this correctly.

F(x) = (1-2x)^2/(3x+1)^3

I proceeded this way:
(3x+1)^3(2)(1-2x) - (1-2x)^2(3)(3x+1)^2(3)=

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

(3x+1)^2(1-2x)[(6x+2)-(9-18x)]

(3x+1)^2(1-2x)(24x-7)

2. Hello, Becky!

Sorry, you made two omissions in the first step . . .

$\displaystyle f(x) \:= \:\frac{(1-2x)^2}{(3x+1)^3}$

I proceeded this way:

$\displaystyle f'(x)\:=\3x+1)^3(2)(1-2x)$(-2) $\displaystyle - (1-2x)^2(3)(3x+1)^2(3)$
. . . . . . . . . . all over denominator-squared

It should have looked like this:
. . $\displaystyle f'(x)\:=\:\frac{(3x+1)^3(2)(1-2x)(-2) - (1-2x)^2(3)(3x+1)^2(3)}{(3x+1)^6}$

. . . . . . $\displaystyle = \;\frac{-4(3x+1)^3(1-2x) - 9(1-2x)^2(3x+1)^2}{(5x+1)^6}$

Factor: . $\displaystyle f'(x)\:=\:\frac{-(3x+1)^2(1-2x)\,\left[4(3x+1) + 9(1-2x)\right]}{(3x+1)^6}$

. . . . . . .$\displaystyle f'(x) \:=\:\frac{-(1-2x)(12x + 4 + 9 - 18x)}{(3x+1)^4}$

. . . . . . .$\displaystyle f'(x)\:=\:\boxed{\frac{-(1-2x)(13 - 6x)}{(3x+1)^4}}\;\text{ or }\;\boxed{\frac{(2x-1)(13-6x)}{(3x+1)^4}}$