Solution
= Solution
Since $f(0)=0$, its <difference quotient> is
$$
\frac{f(h)-f(0)}h
=h\operatorname{sign}(h)\left|\cos\frac\pi h\right|
=|h|\left|\cos\frac\pi h\right|\qquad(h\ne0).
$$
The <absolute value> of <cosine> is at most $1$, so this quotient lies between $0$ and $|h|$. The <squeeze theorem> proves
$$
\boxed{f\text{ is differentiable at }0,\qquad f'(0)=0.}
$$