Find dy/dx for $\displaystyle x^3y^3 - y = x$

Here's what I did;

$\displaystyle 3x^2 3y^2 y' - y' = 1$

$\displaystyle y'(3x^2 3y^2 - 1) = 1$

$\displaystyle y' = 1 / (3x^2 3y^2 - 1)$

Could someone point out to me where I went wrong? The given answer is $\displaystyle 1 - 3x^2 y^3 / (3x^3 y^2 - 1)$