Answer
$$y'=\sin\frac{1}{x}-\frac{1}{x}\cos\frac{1}{x}$$
Work Step by Step
$$y'=(x\sin\frac{1}{x})'=x'\sin\frac{1}{x}+x(\sin\frac{1}{x})'=1\cdot\sin\frac{1}{x}+x\cdot\cos\frac{1}{x}(\frac{1}{x})'=
\sin\frac{1}{x}+x\cos\frac{1}{x}(-\frac{1}{x^2})=\sin\frac{1}{x}-\frac{1}{x}\cos\frac{1}{x}$$
