Answer
$$f'\left( x \right) = 3\left( 3 \right){\left( {3x - 1} \right)^{3 - 1}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {\left( {3x - 1} \right)^3} \cr
& {\text{We need prove that }}f'\left( x \right) = 3\left( 3 \right){\left( {3x - 1} \right)^{3 - 1}} \cr
& {\text{Thus}}{\text{,}} \cr
& {\text{Expanding the binomial}} \cr
& f\left( x \right) = {\left( {3x} \right)^3} - 3{\left( {3{x^2}} \right)^2}\left( 1 \right) + 3\left( {3x} \right){\left( 1 \right)^2} - {\left( 1 \right)^3} \cr
& f\left( x \right) = 27{x^3} - 27{x^2} + 9x - 1 \cr
& {\text{Differentiate by using the rule }}\frac{d}{{dx}}\left[ {{x^r}} \right] = r{x^{r - 1}} \cr
& f'\left( x \right) = 27\left( 2 \right){x^{3 - 1}} - 27\left( 2 \right)x + 9 \cr
& f'\left( x \right) = 81{x^2} - 54x + 9 \cr
& {\text{Factoring}} \cr
& f'\left( x \right) = 9\left( {9{x^2} - 6x + 1} \right) \cr
& f'\left( x \right) = 9{\left( {3x - 1} \right)^2} \cr
& or \cr
& f'\left( x \right) = 3\left( 3 \right){\left( {3x - 1} \right)^{3 - 1}} \cr
& {\text{We have verified that for }} \cr
& f\left( x \right) = {\left( {3x - 1} \right)^3} \Rightarrow f'\left( x \right) = 3\left( 3 \right){\left( {3x - 1} \right)^{3 - 1}} \cr
& f\left( x \right) = {\left( {mx + b} \right)^n}\,\, \Rightarrow \,\,f\left( x \right) = nm{\left( {mx + b} \right)^{n - 1}}\, \cr} $$
