Rewrite #f(x)=(ln(x^2-x^3+x^4))^(-1)#.
Use the chain rule to differentiate from here:
#f'(x)=-(ln(x^2-x^3+x^4))^(-2)d/dx[ln(x^2-x^3+x^4)]#
Find just #d/dx[ln(x^2-x^3+x^4)]#.
Using the chain rule, remembering that #d/dx[ln(x)]=1/x#, recall that #d/dx[ln(u)]=(u')/u#.
#d/dx[ln(x^2-x^3+x^4)]=(d/dx[x^2-x^3+x^4])/(x^2-x^3+x^4)=(2x-3x^2+4x^3)/(x^2-x^3+x^4)#
Plug back in.
#f'(x)=-(ln(x^2-x^3+x^4))^(-2)((2x-3x^2+4x^3)/(x^2-x^3+x^4))#
#f'(x)=-1/(ln^2(x^2-x^3+x^4))((2x-3x^2+4x^3)/(x^2-x^3+x^4))#
#f'(x)=-(x(4x^2-3x+2))/(x(x^3-x^2+x)ln^2(x^4-x^3+x^2))#
#f'(x)=-(4x^2-3x+2)/((x^3-x^2+x)ln^2(x^4-x^3+x^2))#
