Solutions

Notice that $f(0)=0$ since

$$|f(0)|\le |0|^\alpha\implies 0\le f(0)\le 0.$$

Also,

$$0\le \left|\frac{f(x)-f(0)}{x}\right|=\left|\frac{f(x)}{x}\right|\le \frac{|x|^\alpha}{|x|}=|x|^{\alpha-1}$$

and

$$\lim_{x\to 0}|x|^{\alpha-1}=0$$

because $\alpha>1$. Hence,

$$\lim_{x\to 0}\left|\frac{f(x)}{x}\right|=0$$

by the Squeeze Theorem, and consequently

$$f'(0)=\lim_{x\to 0}\frac{f(x)}{x}=0.$$