1. **State the problem:** Find the derivative $f'(x)$ of the function $f(x) = \frac{1}{-3x - 3}$ using the definition of the derivative. 2. **Recall the definition of the derivative:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Apply the definition:** $$f'(x) = \lim_{h \to 0} \frac{\frac{1}{-3(x+h) - 3} - \frac{1}{-3x - 3}}{h}$$ 4. **Simplify the expressions inside the limit:** $$= \lim_{h \to 0} \frac{\frac{1}{-3x - 3h - 3} - \frac{1}{-3x - 3}}{h}$$ 5. **Find a common denominator for the numerator:** $$= \lim_{h \to 0} \frac{\frac{-3x - 3 - (-3x - 3h - 3)}{(-3x - 3h - 3)(-3x - 3)}}{h}$$ 6. **Simplify the numerator's numerator:** $$-3x - 3 - (-3x - 3h - 3) = -3x - 3 + 3x + 3h + 3 = 3h$$ 7. **Rewrite the expression:** $$= \lim_{h \to 0} \frac{\frac{3h}{(-3x - 3h - 3)(-3x - 3)}}{h} = \lim_{h \to 0} \frac{3h}{h \cdot (-3x - 3h - 3)(-3x - 3)}$$ 8. **Cancel $h$ in numerator and denominator:** $$= \lim_{h \to 0} \frac{3}{(-3x - 3h - 3)(-3x - 3)}$$ 9. **Evaluate the limit as $h \to 0$:** $$= \frac{3}{(-3x - 3)(-3x - 3)} = \frac{3}{(-3x - 3)^2}$$ 10. **Final answer:** $$f'(x) = \frac{3}{(-3x - 3)^2}$$ This means the derivative of $f(x)$ is $\frac{3}{(-3x - 3)^2}$, found using the limit definition of the derivative.