Answer
Saddle point at $(1,-1)$; Local minimum at $(0,0)=0$
Work Step by Step
Since, we have $f_x(x,y)=12x-6x^2+6y=0, f_y(x,y)=6y+6x=0$
After solving the above two equations, we get
The critical point is: $(0,0)$ and $(1,-1)$
As per second derivative test, for $(1,-1)$ we have
$D=f_{xx}f_{yy}-f^2_{xy}=-36$ and $D=-36 \lt 0$
Thus, saddle point at $(1,-1)$
Now, for $(0,0)$, we have
$D=f_{xx}f_{yy}-f^2_{xy}=36$ and $D=36 \gt 0$ and $f_{xx}=12 \gt 0$
Thus, we have a local minimum at $f(0,0)=0$
Hence: Saddle point at $(1,-1)$; Local minimum at $(0,0)=0$
