How to prove that a point is not local maximum or local minimum of the function $f(x,y)$?

Question

I was faced with this question:

Given that $f(x,y)=x^3-y^3-3x^2+6y^2+3x-12y+8$.
Prove that $(1,2)$ is a critical point of $f$, and that it's not a local minimum or a local maximum.

It's my first time dealing with a question like this, I have found the point and found that $f_{xx}f_{yy} - (f_{xy})^2 = 0$.
And now I need to prove that it's not maximum or minimum, how do I do that? The only way I thought about is just substituting points until I get $f(x,y) > f(1,2)$ and another point $f(x,y) < f(1,2)$.
But that obviously isn't an efficient way and might be wrong.

I would appreciate any help in how to deal with this question and questions like these in general whenever I get a point that isn't a min/max/saddle, how can I decide what point is it?

Answer 1

$f(x,y)=(x-1)^3-(y-2)^3+1$ and $f(1,2)=1$.

For any $\varepsilon>0$

• $f(1+\varepsilon,2)=1+\varepsilon^3>f(1,2)$.
• $f(1,2+\varepsilon)=1-\varepsilon^3<f(1,2)$.

So "near" $(1,2)$ you always can find points above and below $f(1,2)$.

Answer 2

On the vertical line $x=1$, we have $$ f(x,y)=f(1,y)=-y^3+6y^2-12y+9=-(y-2)^3+1 $$ which is strictly decreasing, hence $f$ does not have a local min or max at $(1,2)$.