$\displaystyle dy/dx=(1+3x^2)/(3y^2-6y)$ with $\displaystyle y(0)=1$

It was easy enough getting it to the solution:

$\displaystyle y^3-3y^2-x-x^3+2=0$.

Now it wants me to find the solution where it's valid. The answer in the back of the book is that $\displaystyle |x|<1$ and the hint was find where the integral curve has a vertical tangent. I'm not really sure how they arrived to this solution though.

Thanks for any help