Answer
Two Saddle points at $(0,1)$ and $(3,-2)$
Local minimum point at $f(3,1)=-34$ and Local maximum point at $f(0,-2)=20$
Work Step by Step
Since, we have $f_x(x,y)=6x^2-18x=0, f_y(x,y)=6y^2+6y-12=0$
After solving the above two equations, we get
The critical point is: $(0,-2),(0,1),(3,-2), (3,1)$
As per the second derivative test, for $(0,-2)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=180$ and $D=180 \gt 0$
and $f_{xx} \gt 0$
Thus, we have: Local minimum point at $f(2,1)=-30$
As per the second derivative test, for $(-2,-1)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=180$ and $D=180 \gt 0$ and $f_{xx} \lt 0$
Thus, we have: Local maximum point at $f(-2,-1)=30$
As per the second derivative test, for $(0,\sqrt 5)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=-180$ and $D=-180 \lt 0$
Thus, we have: Saddle point at $(0,1)$
Now, for $(3,-2)$ , we have
$D=f_{xx}f_{yy}-f^2_{xy}=-180$ and $D=-180 \lt 0$
Hence, we have: two Saddle points at $(0,1)$ and $(3,-2)$
Local minimum point at $f(3,1)=-34$ and Local maximum point at $f(0,-2)=20$
