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We have to prove if the derivative exists at $0$. It's clear that the function is continuous because: $$\lim_ {x \rightarrow 0}x^2\sin\left (\frac {1} {x}\right) = 0\times\lim_ {x \rightarrow 0}\sin\left (\frac {1} {x}\right)$$ and $$x\mapsto\sin\left (\frac {1} {x}\right)$$ is bounded.
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