If $f$ has a relative extremum at $c$, then either $f^{'}(c) = 0$ or $f^{'}(c)$ does not exist. Can one prove the opposite? 
If $f$ has a relative extremum at $c$, then either $f'(c) = 0$ or $f'(c)$ does not exist. Can one prove the opposite?

The statement is due to Fermat. It shows that one can find the relative extrema of $f$ by finding the points $c$ such that $f'(c)=0$ or $f'(c)$ doesn't exist.
However can one show that, if  $f'(c)=0$ or $f'(c)$ does not exist for some value $c$, then $f$ attains a relative extremum at $c$ ?
 A: It's not true.
\begin{align}
f(x)&=x^3\\
f'(0)&=0
\end{align}
However, $0$ is neither a local minima or maxima. It is in fact a saddle point.
You can have points like these which are stationary (i.e. they have $0$ derivative) but they aren't local extrema. 
Roughly speaking, these are points where locally you have increase in one direction and decrease in another. 
A: No, because the other way isn't true.  Here's an example:
$$
f(x) = x^3 \\
f'(x) = 3x^2 \rightarrow f'(0) = 0
$$
But $x = 0$ isn't an extremum.  The only definition for an extremum is that if the derivative's sign changes then you definitely have an extremum.  If the function is differentiable (i.e. the derivative is continuous), then this can only occur when $f'(x) = 0$.  Otherwise, there may be a jump discontinuity in the derivative where it jumps from positive to negative or vice versa (like with $f(x) = |x|$).
edit
I should always be careful about making absolute statements.  If the sign of the derivative changes and the function exists at that point, then it is an extremum.  So for instance, $f(x) = \frac{1}{x^2}$: well the sign of the derivative changes at $x = 0$, but the function doesn't exist at that point and thus is not an extremum (unless you're like me an you like to consider infinities as max's or mins--especially when they're in the middle of a graph).
