Answer
Minimum value: $f(1,\dfrac{1}{2})=-1$
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 $(x,y)=(1,\dfrac{1}{2})$
$D=108 \gt 0$ ; and $f_{xx} =36\gt 0$
Thus, when $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
For $(x,y)=(0,0)$
$D=-36 \lt 0$
When$D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is a not a local minimum and local maximum or, a saddle point.
Now, $f(1,\dfrac{1}{2})=x^3-6xy+8y^3=(1)^3-6(1)(1/2)+8(1/2)^3$
Therefore, we have Minimum value: $f(1,\dfrac{1}{2})=-1$
Saddle point $(0,0)$
