= Solution
A real <function> is <differentiable> at an interior point $x$ when the finite <limit>
$$
f'(x)=\lim_{h\to0,\ h\ne0}\frac{f(x+h)-f(x)}h
$$
exists, with $h$ restricted so $x+h$ lies in its domain. In quantified form, for every $\varepsilon>0$ there is $\delta>0$ such that $0<|h|<\delta$ implies that the <difference quotient> differs from $f'(x)$ by less than $\varepsilon$.
To prove <differentiability implies continuity>, the convergent <difference quotient> is bounded by $|f'(x)|+1$ for sufficiently small nonzero $h$. Thus $|f(x+h)-f(x)|\le(|f'(x)|+1)|h|\to0$. This is exactly <continuity> at $x$.
At zero, the particular function has <difference quotient> $h\sin(1/h)$. The bound $|h\sin(1/h)|\le|h|$ and the <squeeze theorem> give
$$
\boxed{f'(0)=0.}
$$
For $x\ne0$, the <product rule> and <chain rule> give
$$
f'(x)=2x\sin(1/x)-\cos(1/x).
$$
At $x_k=1/(2\pi k)$ this equals $-1$, although $x_k\to0$ and $f'(0)=0$. At $y_k=1/((2k+1)\pi)$ it equals $+1$. Hence the derivative has no limit at zero and is \b[not continuous there]. This demonstrates that <differentiability does not imply continuity of the derivative>, even though the function itself is continuous.
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