From the title, consider the function $f(x) = x + x^2 S(1/x^2)$ if $x\ne 0$ and $f(x) = 0$ if $x=0$.
$S(x) = 1-|x-1|$ if $-1\le x \le 3$ and $S(x) = S(x+4)$ for all real numbers $x$.
I can prove the derivative at 0 is greater than 0 by the Squeeze Theorem but am unsure about the interval piece of the question.
Note that $f$ is differentiable when $S(1/x^2)$ is differentiable or when $x=0$. $S(y)$ is differentiable when $y \neq 1 + 2p$, where $p \in \mathbb Z$. Translating that to $S(1/x^2)$ and then to $f$, we see that $f(x)$ is differentiable when $x\neq \pm \frac{1}{\sqrt{1+2p}}$, where $p =0,1, 2, 3, \cdots$. In that intervals, we have $$\tag{1} f'(x) = 1 + 2x S\left(\frac{1}{x^2}\right) -\frac{2}{x} S'\left(\frac{1}{x^2}\right) $$ For all $h>0$, there is $N\in \mathbb N$ so that $$\pm \frac{1}{\sqrt{1+2n}} \in [-h,h]$$ for all $n\ge N$. In these intervals, we have either $S' = 1$ or $S' = -1$. Put that in (1), using the fact that we have $\frac{2}{x}$ in front of $S'$ which will dominate the other two term as $x\to 0$, then $f'(x)<0$ in some intervals in $[-h,h]$. Thus $f$ is not strictly increasing in $[-h,h]$.
Remark Life will be much easier if you use $\sin\left(\frac{1}{x^2}\right)$ instead of $S\left(\frac{1}{x^2}\right)$.