ok im haveing trouble with this Find The solution (anti derivative) of y'= 1/(3y^2) i tired bringing up the y^2 to the top and making it (y^-2)/3 then i did the dy/dx thing and tired to move the x's and y's to one side but then i still dont get the answer. i think that i am doing something wrong in my first step. do i bring the y^2 up to y^-2?? if its wrong could you tell me what i am doing wonge. the answer is so posed to be the cube root of x + c; where c is a constant; it comes with one of the rules.