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Counterexample: Choose a positive sequence $c_1 > c_1^2 > c_2 > c_2^2 > c_3 > … \to 0.$ On each $[c_n^2, c_n],$ define $f = c_n^2.$ It might be good to draw a picture right bout now.

Define $f$ on each $(c_{n+1},c_n^2)$ by joining the horizontal steps with a line segment. So far we have an increasing, but not strictly increasing, function on the interval $(0,c_1].$ Let's continue this debacle by setting $f(x) = c_1^2, x > c_1,$ and $f(x) = 0, x\le 0.$ So $f$ is increasing on all of $\mathbb R.$

Set $b_n = c_n, a_n =-c_n.$ Then

$$\frac{f(b_n) - f(a_n)}{b_n-a_n} = \frac{c_n^2}{2c_n}= \frac{c_n}{2} \to 0.$$

But

$$\frac{f(c_n^2) - f(0)}{c_n^2-0} = 1$$

for all $n.$ It follows that $f'(0)$ cannot be $0$ (and in fact, doesn't exist).

Now to find a strictly increasing counterexample, consider $f(x) + x^3.$