Take for example $$
f(t) = \begin{cases}
-x^2 &\text{if $x < 0$} \\
x^2 &\text{if $x \geq 0$.}
\end{cases}
$$
For $x<0$ you have $f''(x) = -2$ while for $x > 0$ you have $f''(x) = 2$. $f$ is continuous as $0$, since $\lim_{t\to0^-} f(t) = \lim_{t\to0^+} f(t) = 0$, but since the second-order left-derivative $-2$ is different from the second-order right-derivative $2$ at zero, the second-order derivative doesn't exist there.
For your second question, maybe things are clearer if stated like this
If the second derivative is greater than zero or less than zero at some point $x$, that point cannot be an inflection point
This is quite reasonable - if the second derivative exists and is positive (negative) at some $x$, than the first derivative is continuous at $x$ and strictly increasing (decreasing) around $x$. In both cases, $x$ cannot be an inflection point, since at such a point the first derivative needs to have a local maximum or minimum.
But if the second derivative doesn't exist, then no such reasoning is possible, i.e. for such points you don't know anything about the possible behaviour of the first derivative.
First Derivative Test:
Find all points where $f'(x)=0$ or is undefined, these are called critical points. Any of these could be an extrema. However that's not the only option. The point could also be an "inflection point".
An inflection point $(a, f(a))$, is a point where the derivative $f'(a)=0$, but where $f'(a-\delta)$ and $f'(a+\delta)$, where $\delta$ is a very small number, have the same sign (both $f'(a-\delta)$ & $f'(a+\delta)$ are positive, or both are negative). An inflection point, is neither a maximum or a minimum.
NOTE: This is not the technical definition of an inflection point, but it does provide a method that will work for any continuous function $f: D \subset \Bbb R \to \Bbb R$, assuming the $\delta$ you choose is "small enough".
Second Derivative Test:
Plug in the values you found for $f'(x)=0$ into the second derivative $f''(x)$. If:
$\begin{cases} f''(x)\gt 0 & \text{Then f(x) is a relative minimum} \\
f''(x)\lt 0 & \text{Then f(x) is a relative maximum} \\f''(x)=0 & \text{Then the test is inconclusive} \end{cases}$
So for your example:
$f'(x)=6x^2-4x^3=0$ when $x=0$ or $x=\frac 32$. And $f'(x)$ is a polynomial so it is defined over the entire domain of $f$.
That means that your potential relative maxima and minima are at $(0,f(0))$ and $(\frac 32, f(\frac 32))$.
Let's use the second derivative test:
$f''(x)=12x-12x^2=12x(1-x)=0$ when $x=0$ or $x=1$ and is again defined everywhere. So let's figure out where it's positive and negative by choosing three points $c_1, c_2, c_3$ such that $c_1 \lt 0 \lt c_2 \lt 1 \lt c_3$. How about $c_1=-1$, $c_2=\frac 12$, and $c_3=2$ -- those seem like pretty easy numbers.
Plugging those into $f''(x)$, we get
$f''(-1)=-12-12=-24\lt 0$
$f''(\frac 12)=6-3=3 \gt 0$ and
$f''(2) = 24-48=-24 \lt 0$
So we've found that $f(x)$ is concave up on the interval $(0,1)$ and concave down on the set $(-\infty, 0)\cup(1,\infty)$.
So now let's evaluate the points $x=0$ and $x=\frac 32$ that we found via the first derivative test. We see that $\frac 32 \in (-\infty, 0)\cup(1,\infty)$, so we know that the point $(\frac 32, f(\frac 32))$ is a relative maximum.
On the other hand $x=0$ is inconclusive with the second derivative test. So let's check if it's an inflection point via the method I described above. Let's make our $\delta=\frac 14$ and evaluate $f'(-\frac 14)$ and $f'(\frac 14)$.
$f'(-\frac 14)=6(\frac {1}{16})-4(\frac{-1}{64})=\frac {7}{16} \gt 0$ and
$f'(\frac 14)=6(\frac {1}{16})-4(\frac{1}{64})=\frac {5}{16} \gt 0$
So $(0,f(0))$ is an inflection point of $f(x)$ and thus not an extrema.
Best Answer
If $f(x)$ has a local max or min at $x=0$, then $f'(0)=0$. Now suppose $f(x)$ has an inflection point at $x=0$.
(case 1) $f'(x)$ is increasing on $(-\epsilon,0]$ and decreasing on $[0,\epsilon)$. Then that means $f'(x) \leq 0$ on $(-\epsilon,\epsilon)$. So $f(x)$ is strictly decreasing on $(-\epsilon,\epsilon)$ so there can't possibly be a local max/min at $x=0$.
(case 2) $f'(x)$ is decreasing on $(-\epsilon,0]$ and increasing on $[0,\epsilon)$. Then that means $f'(x) \geq 0$ on $(-\epsilon,\epsilon)$. So $f(x)$ is strictly increasing on $(-\epsilon,\epsilon)$ so there can't possibly be a local max/min at $x=0$.