Answer
$$f'\left( x \right) = \left( {24{{\left( {1 + 2x} \right)}^3} + 6\left( {1 - x} \right)} \right){\left[ {{{\left( {1 + 2x} \right)}^4} - {{\left( {1 - x} \right)}^2}} \right]^2}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left[ {{{\left( {1 + 2x} \right)}^4} - {{\left( {1 - x} \right)}^2}} \right]^3} \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dz}}{\left[ {{{\left( {1 + 2x} \right)}^4} - {{\left( {1 - x} \right)}^2}} \right]^3} \cr
& {\text{Use the chain rule and the power rule }}\frac{d}{{dx}}\left[ {{u^n}} \right] = n{u^{n - 1}}u' \cr
& f'\left( x \right) = 3{\left[ {{{\left( {1 + 2x} \right)}^4} - {{\left( {1 - x} \right)}^2}} \right]^2}\frac{d}{{dz}}\left[ {{{\left( {1 + 2x} \right)}^4} - {{\left( {1 - x} \right)}^2}} \right] \cr
& f'\left( x \right) = 3{\left[ {{{\left( {1 + 2x} \right)}^4} - {{\left( {1 - x} \right)}^2}} \right]^2}\left( {4{{\left( {1 + 2x} \right)}^3}\left( 2 \right) - 2\left( {1 - x} \right)\left( { - 1} \right)} \right) \cr
& {\text{Multiply and simplify}} \cr
& f'\left( x \right) = 3{\left[ {{{\left( {1 + 2x} \right)}^4} - {{\left( {1 - x} \right)}^2}} \right]^2}\left( {8{{\left( {1 + 2x} \right)}^3} + 2\left( {1 - x} \right)} \right) \cr
& f'\left( x \right) = \left( {24{{\left( {1 + 2x} \right)}^3} + 6\left( {1 - x} \right)} \right){\left[ {{{\left( {1 + 2x} \right)}^4} - {{\left( {1 - x} \right)}^2}} \right]^2} \cr} $$
