I don't understand how to prove this considering f(x) is not specifically given....
At zero, $\displaystyle f(0)\leq ||0||^a\implies f(0)=0$. So $\displaystyle \frac{|f(x)-f(0)|}{||x-0||}\leq||x||^{a-1}$.
Take the limit of both sides as $\displaystyle x\to0$ and you get $\displaystyle f'(0)=0$.
If $\displaystyle a=1$, all you know is that $\displaystyle |f'(0)|\leq1$.
Remember that for a function to be diff. at $\displaystyle 0$ there exists a linear transformation $\displaystyle D_f(0)$ such that $\displaystyle \vert \frac{f(h)-f(0)-D_f(0)h}{ \Vert h \Vert } \vert \rightarrow 0$ as $\displaystyle h \rightarrow 0$. Take $\displaystyle D_f(0) \equiv 0$ then $\displaystyle \vert \frac{f(h)-f(0)}{ \Vert h \Vert } \vert \leq \frac{ \Vert h \Vert ^a}{ \Vert h \Vert } = \Vert h \Vert ^{a-1}$ and since $\displaystyle a-1>0$ we have that this goes to $\displaystyle 0$ as $\displaystyle h$ goes to $\displaystyle 0$. This proves that $\displaystyle f$ is diff. at $\displaystyle 0$ and $\displaystyle D_f(0) \equiv 0$. When $\displaystyle a=1$ we can only bound by $\displaystyle 1$, but I can't think of a counter-example at the moment for this case.