$\underset{x\to -1}{\mathop{\lim }}\,\frac{f(x)-f(-1)}{x+1}={f}'(-1)=2$  $y=f(\sin \frac{\pi }{x}-1)\Rightarrow {y}'=(-\frac{\pi }{{{x}^{2}}}\cos \frac{\pi }{x}){f}'(\sin \frac{\pi }{x}-1)$ $\Rightarrow {y}'(1)=(-\frac{\pi }{{{(1)}^{2}}}\cos (\pi ){f}'(\sin \pi -1)=\pi \times {f}'(-1)=2\pi $