Answer
Two Saddle points at $(0,0) , (-2,2)$; Local minimum at $(0,2)=-12$ and Local maximum at $(-2,0)=-4$
Work Step by Step
Since, we have $f_x(x,y)=3x^2+6x=0, f_y(x,y)=3y^2-6y=0$
After solving the above two equations, we get
The critical points are: $(0,0),(0,2),(-2,0), (-2,2)$
As per the second derivative test, for $(0,0)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=-36$ and $D=-36 \lt 0$
Thus, saddle point at $(0,2)$
As per the second derivative test, for $(0,2)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=36$ and $D=36 \gt 0$ and $f_{xx} \gt 0$
Thus, local minimum point at $f(0,2)=-12$
As per the second derivative test, for $(-2,0)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=36$ and $D=36 \gt 0$ and $f_{xx} \lt 0$
Thus, local maximum point at $f(-2,0)=-4$
Now, for $(-2,2)$, we have
$D=f_{xx}f_{yy}-f^2_{xy}=-36$ and $D=-36 \lt 0$
Hence: Two Saddle points at $(0,0) , (-2,2)$; Local minimum at $(0,2)=-12$ and Local maximum at $(-2,0)=-4$
