We'll be using the quotient rule for this one:
#d/dx[f(x)/g(x)]=(f'(x)g(x)-g'(x)f(x))/((g(x))^2)#
Let:
#f(x)=1+ln(3x^2)#
#g(x)=1+ln(4x)#
So:
Use #color(blue)(d/dx[ln(u)]=1/u*u'#
#f'(x)=color(blue)(d/dx[1]+d/dx[ln(3x^2)]=0+1/(3x^2)*d/dx[3x^2]=1/(3x^2)*6x=(6x)/(3x^2)=color(red)(2/x#
#g'(x)=color(blue)(d/dx[1]+d/dx[ln(4x)]=0+1/(4x)*d/dx[4x]=1/(4x)*4=4/(4x)=color(red)(1/x#
By the product rule:
#=(2/x*1+ln(4x)-1/x*1+ln(3x^2))/((1+ln(4x))^2)#
#=((2(1+ln(4x)))/x-((1+ln(3x^2)))/x)/((1+ln(4x))^2#
#=((2(1+ln(4x))-(1+ln(3x^2)))/x)/((1+ln(4x))^2#
Apply the fraction rule for further simplification: #(a/b)/c=a/b*1/c#
#=(2(1+ln(4x))-(1+ln(3x^2)))/x*1/(1+ln(4x))^2#
#=(2(1+ln(4x))-(1+ln(3x^2)))/(x(1+ln(4x))^2)#
We can simplify the numerator if desired
#color(blue)(2(1+ln(4x))-(1+ln(3x^2))=2+2ln(4x)-1-ln(3x^2)#
Applying several log rules...
#color(blue)(2+ln(4x)^2-1-ln(3x^2)#
#color(blue)(1+ln((4x)^2/(3x^2))=1+ln((16x^2)/(3x^2))=color(red)(1+ln(16/3)#
#:.d/dx[(1+ln(3x^2))/(1+ln(4x))]=(1+ln(16/3))/(x(1+ln(4x))^2)#
