y=x^3+3x^2-3x-9. The changes of sign determine the minimum number of turning points. There is therefore at least one (sign changes from + to - along the terms of the polynomial).
y'=3x^2+6x-3=0 at a turning point. This quadratic can be reduced to x^2+2x=1, and completing the square:
x^2+2x+1=2, (x+1)^2=2 so x+1=+sqrt(2), making x=sqrt(2)-1=0.4142 and -(sqrt(2)+1)=-2.4142 the x values of the turning points. The y values are -4(sqrt(2)+1)=-9.657 and 4(sqrt(2)-1)=1.657. So the turning points are (0.4142,-9.657) and (-2.4142,1.657).
Further differentiation tells us what type of turning points we have: y"=6x+6. When x=0.4142 this is positive so the point is minimum; when x=-2.4142 it's negative, so maximum. Therefore, the maximum turning point is (-2.4142,1.657) or (-(sqrt(2)+1),4(sqrt(2)-1)).