Relation between points of inflection and saddle points Let $I$ be an interval and $f\colon I \to \mathbb{R}$ a differentiable function. Suppose the following definitions:
For $x_0 \in I$ the point $(x_0,f(x_0))$ is called saddle point if $f'(x_0) = 0$ but $x_0$ is not a local  extremum of $f$.
For $x_W \in I$ the point $(x_W,f(x_W))$ is called point of inflection if there is a neighborhood $U$ from $x_W$ in $I$ such that $f'$ is strictly monotonic increasing (resp. decreasing) for $x < x_W$ on $U$ and strictly monotinic decreasing (resp. increasing) for $x > x_W$ on $U$.
What is the logical relation between saddle points and points of inflection? 
My first intuitive guess was that a point $(x,f(x))$ is a saddle point iff it is a point of inflection and $f'(x) = 0$. However the implication "$\implies$" seems to be wrong.  Consider the following counterexample:
$$
f(x) = 
\begin{cases} x^4 \cdot \sin\left(\frac{1}{x}\right) & x \neq 0 \\
0 & x = 0
\end{cases}
$$
Then $(0,0)$ is a saddle point but not a point of inflection because the derivativative oscillates on every neighborhood of $0$.
Is this correct so far? Is the other implication true? If so, how to prove it?
 A: Your example does indeed show that a saddle point need not be an inflection point. (The function $x^2\sin(1/x)$ also works, but your example has the virtue of being continuously differentiable.)
In the other direction, if $(a,f(a))$ is a point of inflection and $f'(a) = 0,$ then $(a,f(a))$ is a saddle point. To see this, suppose WLOG that for some small $\delta > 0$ that $f'$ strictly increases in $[a-\delta,a]$ and strictly decreases in $[a,a+\delta].$ In $[a-\delta,a)$ we have $f'(x)<0,$ because these values must be less than $f'(0)=0.$ The same reasoning shows that $f'(x) < 0$ for $x\in (a,a+\delta].$ The mean value theorem then shows $f$ strictly decreases on both $[a-\delta,a]$ and $[a,a+\delta].$ Hence $f$ strictly decreases on $[a-\delta,a+\delta].$ It follows that $f(a)$ is neither a local max. nor min. for $f$ at $a.$
A: OK for the first part.
Let's take a look at the second part (the $\Leftarrow$ implication). It is indeed true.
Let us suppose $(x,f(x))$ is a point of inflexion and $f'(x) = 0$. Then there exists an interval $J = [a,b]$, $x \in J$ and $J \subset I$, where we can suppose by symmetry, that $f'$ is increasing on $[a,x]$ and decreasing on $[x,b]$ (otherwise, we can study $-f$).
If $f' < 0$ on $[a,x[$ and $f' < 0$ on $]x,b]$, we have:
$\forall u \in [a,x[$, $f(u) > f(x)$ and $\forall v \in ]x,b]$,  $f(v) < f(x) $. Consequently, $(x,f(x))$ is not an extremum and is a saddle point.
Then, let us look at the left interval, with a proof by contradiction. If we suppose that there exists $c$ in $[a,x[$ such as $f'(c) >= 0$, as $f'$ is strictly increasing on $[c,x]$, we also have a $d \in [c,x[$ such as $f'(d) > 0$ and $\forall u \in [d,x[$, $f'(u) > f'(d) > 0$. 
By integrating on $[x-h,x]$ with $h \in [0,x-d[$, we have $\forall h \in ]0,x-d[$, $f(x)-f(x-h) > f'(d)h$, then $\frac{f(x-h)-f(x)}{h} <-f'(d) < 0$. Consequently, by taking the limit, $f'(x) \neq 0$, and we have our contradiction.
We can have a similar reasoning for the right interval, and conclude that $f' < 0$ on $[a,x[$, $f'>0$ on $]x,b]$, which means that $(x,f(x))$ is a saddle point.
A: Can only be brief, sorry  you can fill in the gaps. Information available in Wiki.. For z = f(x,y) ; second derivative test. <0 for max, >0 for min, test fails but at saddle points the both signs prevail. E.g., monkey saddle.$ f(x,y) = x^3 - 3 x y^2 $. when considering inflection points along certain directions ( 3 of 6 directions). Like a Col point in mountainous range, one direction upward, one downward, one is neither, topography negative Gauss curvature, inflection along asymptotic direction, normal curvature vanishes..  See also Minimax, Nash equilibrium.
