Answer
Two Saddle points at $(0,\sqrt 5)$ and $(0,-\sqrt 5)$
Local minimum point at $f(2,1)=-30$ and Local maximum point at $f(-2,-1)=30$
Work Step by Step
Since, we have $f_x(x,y)=3x^2+3y^2-15=0, f_y(x,y)=6xy+3y^2-15=0$
After solving the above two equations, we get
The critical point is: $(2,1),(-2,-1),(0,\sqrt 5), (0,-\sqrt 5)$
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 $f_{xx} \gt 0$
Thus, we have a 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 a 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 a saddle point at $(0,\sqrt 5)$
Local maximum point at $f(-2,0)=-4$
Now, for $(0,-\sqrt 5)$ , we have
$D=f_{xx}f_{yy}-f^2_{xy}=-180$ and $D=-180 \lt 0$
Thus, we have a saddle point at $(0,-\sqrt 5)$
Hence, we have two saddle points at $(0,\sqrt 5)$ and $(0,-\sqrt 5)$
We also have a local minimum point at $f(2,1)=-30$ and a local maximum point at $f(-2,-1)=30$
