Answer
Local minimum $f(2,1)=-8$ and saddle point at $(0,0)$
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 not a local minimum and local maximum or, a saddle point.
For $(x,y)=(0,0)$ $D(0,0)=144 \gt 0$ and $f_{xx}(0,0)\lt 0$
For $(x,y)=(2,1)$ $D(2,1)=432 \gt 0$ and $f_{xx}(2,1)=12\gt 0$; as per derivative test, that is, 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.
Also, at points $(2,1)$, $f(2,1)=x^3-12xy+8y^3=(2)^3-12(2)(1)+8(1)^3=-8$
Hence, Local minimum $f(2,1)=-8$ and saddle point at $(0,0)$
