Critical points of surface and type?

Question

If z=x^3 - 3x^2 - 3y^2 _3xy^2 is a surface. Find all the critical points and their type

Answer 1

dz/dx= 3x^2-6x-3y^2=0 (a)
dz/dy= -6y-6xy=0 (b)
from (b)y= 0) and x= -1 which from (a) gives
3x(x-3) =0 which is x=0 and x=3 and the points(0,0) and(3,0)
For x=-1 you get 3+6=3y^2 so y=+-sqrt(3)
(-1,sqrt3) and (-1,-sqrt(3))
So there are four critical points
d2z/dx2= 6x-6
d2z/dy2= -6-6x
d2z/dxdy = -6y
At(0,0) fxx*fyy-fxy^2 = (-6)(-6) -0 = 36 >0 and fxx <0(maximum)
At(3,0) fxx*fyy-fxy^2= 12*(-24)<0 (saddle point)
At(-1 sqrt3) (-12)*0-(6sqrt3)^2<0 saddlde point
At(-1 ,-sqrt3) also a saddle point