Answer
Maximum value: $f(1,0)=f(-1,0) =3$
Saddle point:$(0,0)$
Work Step by Step
Second derivative test: Some noteworthy points to calculate the local minimum, local maximum and saddle point of $f$.
1. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
2.If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
3. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is a not a local minimum and local maximum or, a saddle point.
For $(x,y)=( 1,0)$
$D(1,0)=16 \gt 0$ ; and $f_{xx}(1,0)=-8 \gt 0$
For $(x,y)=( -1,0)$
$D(-1,0)=16 \gt 0$ ; and $f_{xx}(-1,0)=-8 \gt 0$
$D(0,0)=-8 \lt 0$; saddle points
As we are given that $f(x,y)=2-x^4+2x^2-y^2$
$f(1,0)=2-(1)^4+2(1)^2-(0)^2=3$
$f(-1,0)=2-(-1)^4+2(-1)^2-(0)^2=3$
Hence,
Maximum value: $f(1,0)=f(-1,0) =3$
Saddle point:$(0,0)$
