Question 855769
In many baby steps:
{{{(2xy^2)^-2(4x^4y)=(1/(2xy^2)^2)(4x^4y) =4x^4y/(2xy^2)^2
=4x^4y/(2^2x^2(y^2)^2)=4x^4y/4x^2y^4=(4/4)(x^4/x^2)(y/y^4) =1*x^((4-2))(1/y^((4-1)))=1*x^2(1/y^3)=x^2/y^3}}}
or if you like negative exponents {{{x^2y^-3}}}