Answer
$$\frac{{dy}}{{dx}} = \frac{{\left( {6x - 6} \right){{\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]}^2}}}{{\left( {\ln 2} \right)\left( {x - 2} \right)}} + {\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]^3}$$
Work Step by Step
$$\eqalign{
& y = x{\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]^3} \cr
& {\text{differentiate both sides }} \cr
& y' = \left[ {x{{\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]}^3}} \right]' \cr
& {\text{Using product rule}} \cr
& y' = x\left( {{{\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]}^3}} \right)' + {\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]^3}\left( x \right)' \cr
& {\text{By the chain rule}} \cr
& y' = 3x{\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]^2}\left( {{{\log }_2}\left( {{x^2} - 2x} \right)} \right)' + {\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]^3} \cr
& {\text{use }}\left( {{{\log }_a}u} \right) = \frac{{u'}}{{\left( {\ln a} \right)u}} \cr
& y' = 3x{\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]^2}\left( {\frac{{2x - 2}}{{\left( {\ln 2} \right)\left( {{x^2} - 2x} \right)}}} \right) + {\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]^3} \cr
& {\text{simplify}} \cr
& y' = \frac{{\left( {6{x^2} - 6x} \right){{\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]}^2}}}{{\left( {\ln 2} \right)\left( {{x^2} - 2x} \right)}} + {\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]^3} \cr
& y' = \frac{{\left( {6x - 6} \right){{\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]}^2}}}{{\left( {\ln 2} \right)\left( {x - 2} \right)}} + {\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]^3} \cr
& or \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {6x - 6} \right){{\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]}^2}}}{{\left( {\ln 2} \right)\left( {x - 2} \right)}} + {\left[ {{{\log }_2}\left( {{x^2} - 2x} \right)} \right]^3} \cr} $$
