Question 935979
{{{(8x^-3*y^6)^(-1/3) (2x^-3*y^-1)}}}


{{{(1/root(3,(8x^-3*y^6))) (2x^-3*y^-1)}}}


{{{(1/root(3,(8(1/x^3)*y^6))) (2(1/x^3)*(1/y))}}}


{{{(1/root(3,(8y^6/x^3))) (2(1/x^3)*(1/y))}}}



{{{(1/root(3,(2^3*(y^2)^3)/x^3)) (2/(yx^3))}}}


{{{(1/(2y^2/x)) (2/(yx^3))}}}


{{{(cross(x)1/(cross(2)y^2)) (cross(2)1/(yx^cross(3)^2))}}}


{{{(1/y^2) (1/yx^2)}}}


{{{ 1/(y^3x^2)}}} <= your answer

as I can see, option 3) has {{{x^2*y^3}}} and need to be {{{ 1/(y^3x^2)}}}