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$f'(0)=\lim_{x\to0}\dfrac{f(x)-f(0)}{x}$
$=\lim_{x\to0}\dfrac{x^2\sin(1/x)}{x}$
$=\lim_{x\to0}x\sin(\frac{1}{x})=0$
For $x\ne0$,
$f'(x)=2x\sin\left(\dfrac{1}{x}\right)-\cos\left(\dfrac{1}{x}\right)$
Answer: $\boxed{f'(0)=0}$ ✅
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