Value of mixed partial derivatives at critical points.

Question

Suppose $f(x,y):\mathbb{R}^2\rightarrow\mathbb{R}$ is smooth, and $$\left.\frac{\partial f}{\partial x}\right\vert_{(x_0,y_0)}=0\;,\qquad \left.\frac{\partial f}{\partial y}\right\vert_{(x_0,y_0)}=0$$ Can I claim: $$\left.\frac{\partial^2f}{\partial x\partial y}\right|_{(x_0,y_0)}=0$$ My intuition is that this is valid, and I have yet to find a conter-example.

Here is my intuition:
At extremas, i.e minimas and maximas, the gradient is 0. Taking a small step in any direction from the extrema (say in the x direction), the direction of the maximum slope will be either towards or away from the extremum (depending on whether it is either a maxima or minima). Then the slope in the direction perpendicular to that (y direction) is 0, hence $\left.\frac{\partial^2f}{\partial x\partial y}\right|_{(x_0,y_0)}=0$
I think the same logic is valid for saddle points as well, but I'm not sure.
Thanks in advance!
Edit: A counter-example has been given in the comments for this question (a saddle-point), now my question is basically if there is anything wrong with my logic for extremas.

Answer 1

Nope. Try $f(x,y) = x^2+y^2+ xy$. This still has a minimum at the origin (and no other critical point).

The mixed partial derivative is very subtle, but your argument “then the slope …” just doesn’t hold water. At $(h,k)$, $\partial f/\partial x = 2h+k$, so at $(h,0)$ we have $\partial^2 f/\partial y\partial x = \partial/\partial k\,(2h+k) = 1$, not $0$.

Answer 2

You're misunderstanding what the slope in a direction means. In general the slope of $f$ in a direction $v$ at $a$ is given by the directional derivative

$$ \lim_{h \to 0} \frac{f(a + hv)-f(a)}{h}$$

You claim that at a maximum or minimum the slope along the $y$ direction is equal to $0$. This is correct but the slope along the $y$ direction is simply

$$ \frac{\partial f}{\partial y}$$ and not $$ \frac{\partial^2 f}{\partial y \partial x}.$$