1. **State the problem:** Find the limit $$\lim_{h \to 0} \frac{f(z_0+h) - f(z_0)}{h}$$ where $$f(z) = \frac{2z - 1}{3z + 2}$$. 2. **Recognize the limit:** This limit is the definition of the derivative of $$f$$ at $$z_0$$, i.e., $$f'(z_0)$$. 3. **Recall the derivative formula for a quotient:** If $$f(z) = \frac{u(z)}{v(z)}$$, then $$$f'(z) = \frac{u'(z)v(z) - u(z)v'(z)}{(v(z))^2}$$$ 4. **Identify $$u(z)$$ and $$v(z)$$:** $$u(z) = 2z - 1$$ $$v(z) = 3z + 2$$ 5. **Compute derivatives:** $$u'(z) = 2$$ $$v'(z) = 3$$ 6. **Apply the quotient rule:** $$f'(z) = \frac{2(3z + 2) - (2z - 1)(3)}{(3z + 2)^2}$$ 7. **Simplify numerator:** $$2(3z + 2) = 6z + 4$$ $$(2z - 1)(3) = 6z - 3$$ So numerator is: $$6z + 4 - (6z - 3) = 6z + 4 - 6z + 3 = 7$$ 8. **Final derivative expression:** $$f'(z) = \frac{7}{(3z + 2)^2}$$ 9. **Evaluate at $$z_0$$:** $$f'(z_0) = \frac{7}{(3z_0 + 2)^2}$$ **Answer:** $$\boxed{\frac{7}{(3z_0 + 2)^2}}$$