The function $f:\Bbb{R}^2\to\Bbb{R}$ has continuous partial derivatives of at least order $4$. And $$f_x=f_y=f_{xx}=f_{yy}=f_{xy}=0$$ at $(0,0)$.
Also, at least one partial derivatives of order $3$ does not vanish at $(0,0)$.
Now how to show that $f$ can have neither a local maximum nor a local minimum at this critical point.
My attempt: I try to use the Taylor expansion, just like what we did when we introduced the Hessian matrix. However, the case of second order is totally different with the case of third order.
By Taylor's theorem you have $$f(x,y)=f(0,0)+p(x,y)+o\bigl(r^3\bigr)\qquad\bigl(r:=\sqrt{x^2+y^2}\to0\bigr)\ ,\tag{1}$$ whereby $p$ is a homogeneous polynomial of degree $3$ (hence odd) that does not vanish identically. It follows that $p$ has a positive maximum $\mu>0$ and a negative minimum $-\mu<0$ on the unit circle $S^1$. Using $(1)$ you can then easily show that $f(x,y)-f(0,0)$ assumes in any neighborhood of $(0,0)$ strictly positive as well as negative values.