1. **State the problem:** Find the third derivative $\frac{d^3y}{dx^3}$ of the function $$y = e^{-x}(\cos 2x + \sin 2x).$$ 2. **Recall the product rule and chain rule:** For derivatives of products, use $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).$$ 3. **Define components:** Let $$u = e^{-x}, \quad v = \cos 2x + \sin 2x.$$ 4. **Compute first derivatives:** $$u' = -e^{-x}, \quad v' = -2\sin 2x + 2\cos 2x.$$ 5. **First derivative:** $$y' = u'v + uv' = -e^{-x}(\cos 2x + \sin 2x) + e^{-x}(-2\sin 2x + 2\cos 2x) = e^{-x}(-\cos 2x - \sin 2x - 2\sin 2x + 2\cos 2x).$$ Simplify inside parentheses: $$-\cos 2x + 2\cos 2x = \cos 2x, \quad -\sin 2x - 2\sin 2x = -3\sin 2x,$$ so $$y' = e^{-x}(\cos 2x - 3\sin 2x).$$ 6. **Second derivative:** Differentiate $y'$: $$y'' = \frac{d}{dx}[e^{-x}(\cos 2x - 3\sin 2x)] = u'v + uv',$$ where now $$u = e^{-x}, \quad v = \cos 2x - 3\sin 2x,$$ so $$u' = -e^{-x}, \quad v' = -2\sin 2x - 6\cos 2x.$$ Then $$y'' = -e^{-x}(\cos 2x - 3\sin 2x) + e^{-x}(-2\sin 2x - 6\cos 2x) = e^{-x}(-\cos 2x + 3\sin 2x - 2\sin 2x - 6\cos 2x).$$ Simplify: $$-\cos 2x - 6\cos 2x = -7\cos 2x, \quad 3\sin 2x - 2\sin 2x = \sin 2x,$$ so $$y'' = e^{-x}(-7\cos 2x + \sin 2x).$$ 7. **Third derivative:** Differentiate $y''$: $$y''' = \frac{d}{dx}[e^{-x}(-7\cos 2x + \sin 2x)] = u'v + uv',$$ where $$u = e^{-x}, \quad v = -7\cos 2x + \sin 2x,$$ so $$u' = -e^{-x}, \quad v' = 14\sin 2x + 2\cos 2x.$$ Then $$y''' = -e^{-x}(-7\cos 2x + \sin 2x) + e^{-x}(14\sin 2x + 2\cos 2x) = e^{-x}(7\cos 2x - \sin 2x + 14\sin 2x + 2\cos 2x).$$ Simplify: $$7\cos 2x + 2\cos 2x = 9\cos 2x, \quad -\sin 2x + 14\sin 2x = 13\sin 2x,$$ so $$\boxed{\frac{d^3y}{dx^3} = y''' = e^{-x}(9\cos 2x + 13\sin 2x)}.$$