Answer
Local minimum at $(1,1)$;
Saddle point at $(0,0)$
Work Step by Step
Given: $f(x,y)=4+x^3+y^3-3xy$
This gives
$f_x(x,y)=3x^2-3y$ and $f_y(x,y)=3y^2-3x$
and $(x,y)=(0,0)$ and $(1,1)$.
To solve this problem we will take the help of Second derivative test that suggests the following conditions to determine the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$.
i) 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.
ii) 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.
iii) 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 $(x,y)=(0,0)$, we have $D(0,0)=-9 \lt 0$
and For $(x,y)=(1,1)$, we have $D(1,1)=27 \gt 0$ and $f_{xx}(1,1)=6 \gt 0$
Hence, Local minimum at $(1,1)$; Saddle point at $(0,0)$
