Analysis of the problem 
Let $ M $ be the set of all functions $ f $ twice differentiable on $ [0,1] $ such that $ f (0) = f (1) = 0 $ and $ | f '' (x) | \leq 100 $. Find $ \max\limits_ {f \in M} \max\limits_ {x \in [0,1]} | f '(x) | $

I think that the condition of this problem is equivalent to the fact that we maximize the $\sup$ norm of the derivative f from this set. As a maximizing function, we can take a polynomial of the second degree with roots at $0$ and $1$
I got an answer of 50, if I said everything correctly and decided
I took it turns out a polynomial of the second degree, in which the roots are $0$ and $1$, then $P (x) = ax (x-1)$, $a> 0$. Now $| P "(x)| = 2, a  \leq 100$. Take as $a^* = 50$. With this $a^*$ the maximum of the second derivative of P will be achieved, that we and it is necessary. So the maximizing function is found. It's just $50x (x-1)$. The norm of the derivative is $50$
 A: This is a nice puzzle at the calculus level.  Since nobody seems interested in helping here is my advice.
First of all your answer is perfect for half of the problem, but missing a key half.  You show that this maximum is at least as big as 50.  You don't prove that it cannot be any larger.
Here is a sketch that you can flesh out with appropriate computation.
Preliminaries.   If $f\in M$ first note that $f(0)=f(1)$ implies there is a number
$c$ so that  $f'(c)=0$.  Moreover, this also implies that
$$0 = f(1)-f(0) = \int_0^1 f'(t)\,dt  = \int_0^c f'(t)\,dt + \int_c^1 f'(t)\,dt $$
and hence
$$\int_0^c f'(t)\,dt = - \int_c^1 f'(t)\,dt \tag{1}.$$
Step 1.  Consider the case  $c=\frac12$.  The condition $|f''(x)|\leq 100$ and the mean-value theorem applied on $[0,\frac12]$ and on $[\frac12,1]$ shows that
$|f'(x)|\leq 50$ for all values of $x$ in $[0,1]$.
But your most excellent example $p(x)=50x(1-x)$ has $p'(x) = 100x -50$, $p''(x)=100$, $p'(0)=-50$, and $p'(1)=50$.
Thus the value $50$ is attained by at least one function.
Step 2.  Consider the case $c> \frac12$.  By the mean-value theorem as before we must have $|f'(x)|<50$ for all $x$ in $[c,1]$.
But if there is a value $x_0$ in $[0,c]$ with $f'(x_0)> 50$  it must be the case,
since $|f''(x)|\leq 100$,   that
$$ \int_0^c f'(x) > \left|\int_c^1 f'(x)\right|.$$
This contradicts (1) above.
Step 3.  The case $c< \frac12$ is simiilar.
Thus no value of $|f'(x)|$ larger than $50$ can  occur if $c\not=\frac12$. We have already checked that the value $50$ is the largest possible for the case $c=\frac12$  and is actually attained for the function $p(x)$ given.
