The question is to disprove the following:
Be $f$ a continuously differentiable function that maps from $\mathbb{R}\rightarrow\mathbb{R}$ and $f(x) =\mathcal{O}(x^2) $ for $x\rightarrow0$, then it follows that $f'(x) =\mathcal{O}(x)$ for $x\rightarrow0$.
I am searching for a counter-example, but i am having a hard time finding one. Basically we need a function $f(x)$ with the following properties for it to be a counter-example:
$$ \frac{f(x)}{x^2} \leq c, \ \mathrm{but \ not}\ \frac{f'(x)}{x} \leq q $$ for some constants $c,q \geq 0$ and $x\rightarrow0$. So our first fraction must be bounded by some $c$, but the second has to be unbounded.
Polynomials cant work (because of the derivation rule for them) and so far i am out of luck with other functions too.
I found a solution. Take the function $$ f(x) = \begin{cases} x^2\sin({\frac{1}{\sqrt{x}}}), & \text{for $x \neq 0$,} \\ 0, & \text{for $x = 0$.} \end{cases} $$
The function is continuous, because $ -1 \leq\sin(x) \leq 1$ and so for $x \rightarrow0$ the function will go to $0$. It is also differentiable everywhere with choosing $f'(0) = 0$. Finally $f$ is in $\mathcal{O}(x^2)$, because: $$|\frac{x^2\sin(\frac{1}{\sqrt{x}})}{x^2}| =|\sin(\frac{1}{\sqrt{x}})| \leq1,$$and computing $f'$ gives $$f'(x)= 2x\sin(\frac{1}{\sqrt{x}}) - \frac{1}{2}\sqrt{x}\cos(\frac{1}{\sqrt{x}}),$$ and we choose $f'(0)= 0$ as written above. Therefore the derivative is continuous, as both $\sin(x)$ and $\cos(x)$ are bounded by $1$ and $-1$ respectively and the $2x$ and $\sqrt{x}$ will go to $0$. Finally we check $f'(x) = \mathcal{O}(x)$ as following: $$|\frac{2x\sin(\frac{1}{\sqrt{x}})}{x} - \frac{\frac{1}{2}\sqrt{x}\cos(\frac{1}{\sqrt{x}})}{x}| = |2\sin(\frac{1}{\sqrt{x}}) -(\frac{1}{2}\cos(\frac{1}{\sqrt{x}}))\cdot\frac{1}{\sqrt{x}}| =(*),$$ and since both $\sin(x)$ and $\cos(x)$ will be bounded, the $\frac{1}{\sqrt{x}}$ part will decide the behavior when taking the limit. Therefore $(*) \rightarrow\infty $ for $x \rightarrow0$, that means $(*)$ is unbounded for $x \rightarrow0$ and with that we get $f'(x) \neq \mathcal{O}(x)$.