Answer
Saddle point at $(0,0)$; Local maximum at $(-1,-1)=1$
Work Step by Step
Since, we have $f_x(x,y)=3x^2+3y=0, f_y(x,y)=3x+3y^2=0$
After solving the above two equations, we get
The critical point is: $(0,0)$ and $(-1,-1)$
As per second derivative test, for $(0,0)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=-9$ and $D=-9 \lt 0$
Thus, saddle point at $(0,0)$
Now, for $(-1,1)$, we have
$D=f_{xx}f_{yy}-f^2_{xy}=27$ and $D=27 \gt 0$ and $f_{xx}=-6 \lt 0$
Thus, local maximum at $(-1,-1)=1$
Hence: Saddle point at $(0,0)$; Local maximum at $(-1,-1)=1$
