1. **State the problem:** We are given the function $$f(x) = -2\sqrt{x^3} - \sqrt{x}$$ and need to find its derivative at $$x=3$$, i.e., $$f'(3)$$. 2. **Rewrite the function using exponents:** Recall that $$\sqrt{x} = x^{1/2}$$ and $$\sqrt{x^3} = (x^3)^{1/2} = x^{3/2}$$. So, $$f(x) = -2x^{3/2} - x^{1/2}$$. 3. **Differentiate term-by-term:** Using the power rule $$\frac{d}{dx} x^n = nx^{n-1}$$, $$f'(x) = -2 \cdot \frac{3}{2} x^{3/2 - 1} - \frac{1}{2} x^{1/2 - 1} = -3x^{1/2} - \frac{1}{2} x^{-1/2}$$. 4. **Simplify the derivative:** $$f'(x) = -3\sqrt{x} - \frac{1}{2\sqrt{x}}$$. 5. **Evaluate at $$x=3$$:** $$f'(3) = -3\sqrt{3} - \frac{1}{2\sqrt{3}}$$. 6. **Express as a single fraction:** Find common denominator $$2\sqrt{3}$$: $$f'(3) = \frac{-3\sqrt{3} \cdot 2\sqrt{3}}{2\sqrt{3}} - \frac{1}{2\sqrt{3}} = \frac{-6 \cdot 3 - 1}{2\sqrt{3}} = \frac{-18 - 1}{2\sqrt{3}} = \frac{-19}{2\sqrt{3}}$$. 7. **Rationalize the denominator:** Multiply numerator and denominator by $$\sqrt{3}$$: $$f'(3) = \frac{-19 \sqrt{3}}{2 \cdot 3} = \frac{-19 \sqrt{3}}{6}$$. **Final answer:** $$f'(3) = \frac{-19 \sqrt{3}}{6}$$.