why does contour lines intersect at saddle points? Consider functions of two variables. A saddle point of $f(x,y)$ is a point in the domain of $f$ (or on the graph of $f$ by an abuse of language) where the gradient is $\bf 0$, but which is not a local extremum of the function. A classic example is $(0,0)$ for $f(x,y)=y^2-x^2$. However, there are two types of saddle points.
Type I: the graph looks like a saddle.
Type II: the graph does not look like a saddle. e.g. $f(x,y)=x^3$, and $f(x,y)=x^2+y^3$.
I have two questions:
Question 1: How to rigorously define saddle points of the above two types?
Question 2: A folklore theorem says $P$ is a saddle point of type I if and only if the contour lines (aka level curves) intersect at $P$. I don't find a proof of such a result, and I'm not sure if the theorem holds as "if and only if" or just holds in one direction.
 A: A reasonable definition for saddle points of type I is the following:

Let $f\colon\mathbf{R}^2\to\mathbf{R}$ be a smooth function, a saddle point of type I of $f$ is a point $p\in\mathbf{R}^2$ such that $\frac{\partial f}{\partial x}(p)=0=\frac{\partial f}{\partial y}(p)$ and $\begin{bmatrix}\frac{\partial^2 f}{\partial x^2}(p) & \frac{\partial^2 f}{\partial x\partial y2}(p)\\\frac{\partial^2 f}{\partial y\partial x}(p) & \frac{\partial^2 f}{\partial y^2}(p)\end{bmatrix}$ has nonpositive determinant.

Regarding the saddle points of type II, you can always say that the previous matrix has zero determinant (it cannot have nonnegative determinant, otherwise $p$ would be a maximum or a minimum). However, from the differential topology view point, this is not such a great definition, since the level sets around such critical points can be very different looking.
Using a standard result called Morse's lemma, around a saddle point of type I, there exist local coordinates such that $f(x,y)=x^2-y^2$, in this local model, it is clear that the level curves $f^{-1}(0)$ intersects at $(0,0)$. The reverse implication is false, a counter example is given by $f(x,y)=x^3-3xy^2$ for which $(0,0)$ is a saddle point of type II, even though $f^{1}(0)$ consists of three transversally intersecting curves.
