= Solution
The <difference quotient> at zero is
$$
\frac{f(h)-f(0)}h=\frac{|h|^p}{h}\cos(1/h),\qquad h\ne0.
$$
If $p>1$, its <absolute value> is at most $|h|^{p-1}\to0$, so $f$ is <differentiable> at zero with <derivative> $0$.
To exclude $p\leq1$, use $h_n=1/(2\pi n)$ and $k_n=1/(2\pi n+\pi/2)$. The <difference quotient> equals $(2\pi n)^{1-p}$ along the first <sequence>, and $0$ along the second. For $p=1$ the first value is $1$, while for $p<1$ it is unbounded. Hence a common finite <limit> is impossible. Therefore
$$
\boxed{f\text{ is differentiable at }0\iff p>1,\qquad f'(0)=0\text{ when }p>1.}
$$
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