If $f$ is a strictly increasing function, shouldn't $f'$ be always positive and never zero? Apparently there's this situation where the derivative can be $0$ if it's only at discrete points and not an interval.
How is that possible, when strictly increasing functions are, by definition, never of a zero slope?
On
A function $f : \mathbb R \to \mathbb R$ is strictly increasing if, for every $b > a$, it follows that $f(b) > f(a)$. This of course is satisfied by $f(x) = x^3$, since if $b > a$ implies $$f(b) - f(a) = b^3 - a^3 = (b - a)(a^2 + ab + b^2) = \frac{1}{4}(b-a)\left((b-a)^2 + 3(b+a)^2\right).$$ Since $b-a > 0$ and no square is negative, all factors are positive, thus $f(b) - f(a) > 0$.
However, $f'(x) = 3x^2$ so $f'(0) = 0$.
"strictly increasing functions are, by definition, never of a zero slope": that is not the definition of a strictly increasing function. The correct one says
$$x_0<x_1\implies f(x_0)<f(x_1).$$
Geometrically, this can be expressed as "the slope of any chord is positive", but not as "the slope of any tangent is positive".