Find dy/dx for the following:

y^3=x^3/(2-xy)

This is my work.. can someone check to see if I have the correct answer and method? Thanks

d/dx (3y^2)=(3x^2)(2-xy)-x^3[(-1)y+x(1)dy/dx]/(2-xy)^2

d/dx (3y^2)=(3x^2)(2-xy)+x^3y-x(dy/dx)]/(2-xy)^2

(2-xy)^2(3y^2)(dy/dx)+x (dy/dx)=3x^2(2-xy)+x^3y

dy/dx[(2-xy)^2(3y^2)+x]=3x^2(2-xy)+x^3y

dy/dx=[3x^2(2-xy)+x^3y]/[(2-xy)^2(3y^2)+x]