How do I find the vertex?
I set y to zero then I get -3? I think.
No, you do NOT set y equal to 0. That would give the y-intercept which, in general, has nothing to with the vertex.
Since a square is never 0, $\displaystyle x= 3(y+1)^2- 3$, is never less than -3. It is -3 when the square is 0- that is, when y+ 1= 0.