Question 289628
Use a substitution and the chain rule. 
Let 
{{{u=4x^3-x^2}}}
{{{y=ln(u)}}}
From the chain rule, 
{{{ dy/dx=dy/du*du/dx}}}

{{{ dy/du=1/u=1/(4x^3-x^2)=1/(x^2(4x-1))}}}

{{{ du/dx=12x^2-2x=2x(6x-1)}}}

So then
{{{ dy/dx=1/(x^2(4x-1))*(2x(6x-1)) }}}
{{{ dy/dx=2(6x-1)/(x(4x-1)) }}}