1. **Problem:** Find the derivative of $y = \frac{\sin x}{\cos x - \sin x}$. 2. **Formula:** Use the quotient rule for derivatives: $$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ where $u = \sin x$ and $v = \cos x - \sin x$. 3. **Calculate derivatives:** $\frac{du}{dx} = \cos x$ $\frac{dv}{dx} = -\sin x - \cos x$ 4. **Apply quotient rule:** $$\frac{dy}{dx} = \frac{(\cos x - \sin x)(\cos x) - (\sin x)(-\sin x - \cos x)}{(\cos x - \sin x)^2}$$ 5. **Simplify numerator:** $$= (\cos x)(\cos x) - (\sin x)(\cos x) + (\sin x)(\sin x) + (\sin x)(\cos x)$$ 6. **Combine like terms:** $$= \cos^2 x - \sin x \cos x + \sin^2 x + \sin x \cos x$$ 7. **Cancel terms:** $$- \sin x \cos x + \sin x \cos x = \cancel{- \sin x \cos x} + \cancel{\sin x \cos x} = 0$$ 8. **Resulting numerator:** $$= \cos^2 x + \sin^2 x$$ 9. **Use Pythagorean identity:** $$\cos^2 x + \sin^2 x = 1$$ 10. **Final derivative:** $$\frac{dy}{dx} = \frac{1}{(\cos x - \sin x)^2}$$ **Answer:** $$\boxed{\frac{dy}{dx} = \frac{1}{(\cos x - \sin x)^2}}$$