Answer
Saddle point at $(0,0)$
Local minimum point at $f(-1,1)=-2$ and Local minimum point at $f(1,-1)=-2$
Work Step by Step
Since, we have $f_x(x,y)=4x^3+4y=0, f_y(x,y)=4y^3+4x=0$
After solving the above two equations, we get
The critical point is: $(0,0),(1,-1),(-1,1)$
As per the second derivative test, for $(0,0)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=-16$ and $D=-16 \lt 0$
Thus, we have: Saddle points at $(0,0)$
As per the second derivative test, for $(1,-1)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=128$ and $D=128 \gt 0$ and $f_{xx} \gt 0$
Thus, we have: Local minimum point at $f(1,-1)=-2$
As per the second derivative test, for $(-1,1)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=128$ and $D=128 \gt 0$ and $f_{xx} \gt 0$
Thus, we have: Local minimum point at $f(-1,1)=-2$
Hence, we have: Saddle point at $(0,0)$
Local minimum point at $f(-1,1)=-2$ and Local minimum point at $f(1,-1)=-2$
