I assume we are differentiating 4/(4x-5^(3x)), which we can call f(x).
Break this down a bit. To differentiate 5^(3x) we need to replace 5 by e^(ln(5)), then the expression becomes e^(3xln(5)), which is differentiable: 3ln(5)e^(3xln(5))=3ln(5)(5^(3x)).
Now we put u=4x-5^(3x), du/dx=4-3ln(5)(5^(3x)).
f(u)=4/u=4u⁻¹, df/dx=df/du×du/dx=-4u⁻²(4-3ln(5)(5^(3x)).
f'(x)=-4(4-3ln(5)(5^(3x))/(4x-5^(3x))².
Note also that 5^(3x)=125^x.