find $\displaystyle dy/dx$ by implicit differentiation

$\displaystyle \tan^3{(xy^2+y)} = 1$

didn't know how to deal with the $\displaystyle \tan^3$

the answer is:

$\displaystyle \frac

{1-3y^2\tan^2(xy^2+y)\sec^2(xy^2+y)}

{3(2xy+1)tan^2(xy^2+y)sec^2(xy^2+y)}$