Answer
$$\frac{{dy}}{{dx}} = - \frac{{3\sin \left( {\ln \left| {2{x^3}} \right|} \right)}}{x}$$
Work Step by Step
$$\eqalign{
& y = \cos \left( {\ln \left| {2{x^3}} \right|} \right) \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = {D_x}\left[ {\cos \left( {\ln \left| {2{x^3}} \right|} \right)} \right] \cr
& {\text{use the chain rule }}{D_x}\left( {\cos u} \right) = - \sin u \cdot {D_x}\left( u \right).{\text{ consider }}u = \ln \left| {2{x^3}} \right| \cr
& \frac{{dy}}{{dx}} = - \sin \left( {\ln \left| {2{x^3}} \right|} \right) \cdot {D_x}\left( {\ln \left| {2{x^3}} \right|} \right) \cr
& {\text{use the chain rule }}{D_x}\left( {\ln u} \right) = \frac{1}{u} \cdot {D_x}\left( u \right) \cr
& \frac{{dy}}{{dx}} = - \sin \left( {\ln \left| {2{x^3}} \right|} \right)\left( {\frac{1}{{2{x^3}}}} \right){D_x}\left( {2{x^3}} \right) \cr
& {\text{solve the derivative and simplify}} \cr
& \frac{{dy}}{{dx}} = - \sin \left( {\ln \left| {2{x^3}} \right|} \right)\left( {\frac{1}{{2{x^3}}}} \right)\left( {6{x^2}} \right) \cr
& \frac{{dy}}{{dx}} = - \sin \left( {\ln \left| {2{x^3}} \right|} \right)\left( {\frac{3}{x}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{{3\sin \left( {\ln \left| {2{x^3}} \right|} \right)}}{x} \cr} $$