= Solution
For $t\ne0$, ordinary differentiation of the two smooth components gives
$$
\boxed{f'(t)=\bigl(2t\sin(1/t)-\cos(1/t),\ 2t\cos(1/t)+\sin(1/t)\bigr)}.
$$
At zero, use the definition of the <derivative> in the <Euclidean norm>:
$$
\left\|\frac{f(h)-f(0)}h\right\|=\|h(\sin(1/h),\cos(1/h))\|=|h|\longrightarrow0.
$$
Thus $f$ has <differentiability> everywhere and $\boxed{f'(0)=(0,0)}$. The two vectors $(\sin(1/t),\cos(1/t))$ and $(-\cos(1/t),\sin(1/t))$ are orthogonal unit vectors, so the squared <norm> of the displayed derivative is $4t^2+1$. Consequently
$$
\boxed{\|f'(a)-f'(0)\|=\sqrt{1+4a^2}>1\qquad(a\ne0)}.
$$
This <discontinuous derivative of a differentiable plane curve> also shows why componentwise scalar <mean value theorem> conclusions do not supply a single intermediate point reproducing an entire vector-valued difference quotient.
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