Rewrite the terms using exponent notation.

$$\frac{dy}{dx} = 2\left(-\frac{1}{2}\right)x^{-1/2 - 1} + \frac{2}{3}x^{2/3 - 1} - (-1)x^{-1 - 1}$$ $$\frac{dy}{dx} = -x^{-3/2} + \frac{2}{3}x^{-1/3} + x^{-2}$$

$$\frac{dy}{dx} = -\frac{1}{x^{3/2}} + \frac{2}{3x^{1/3}} + \frac{1}{x^2}$$ $$\frac{dy}{dx} = -\frac{1}{x\sqrt{x}} + \frac{2}{3\sqrt[3]{x}} + \frac{1}{x^2}$$

The final answer is $\boxed{-\frac{1}{x\sqrt{x}} + \frac{2}{3\sqrt[3]{x}} + \frac{1}{x^2}}$.

2) Given $y = 3x^4 + (2xp)^0 - 3q + \frac{16}{\sqrt[4]{x^3}}$

$(2xp)^0 = 1$ (assuming $2xp \neq 0$), $3q$ is a constant, and $\frac{16}{\sqrt[4]{x^3}} = 16x^{-3/4}$.

$$y = 3x^4 + 1 - 3q + 16x^{-3/4}$$

The derivative of a constant is 0.

$$\frac{dy}{dx} = 3(4)x^{4-1} + 0 - 0 + 16\left(-\frac{3}{4}\right)x^{-3/4 - 1}$$ $$\frac{dy}{dx} = 12x^3 - 12x^{-7/4}$$

$$\frac{dy}{dx} = 12x^3 - \frac{12}{x^{7/4}}$$ $$\frac{dy}{dx} = 12x^3 - \frac{12}{x\sqrt[4]{x^3}}$$

The final answer is $\boxed{12x^3 - \frac{12}{x\sqrt[4]{x^3}}}$.

3) Given $y = \frac{4x^2 + 3x + 2}{x^3}$

$$y = \frac{4x^2}{x^3} + \frac{3x}{x^3} + \frac{2}{x^3}$$ $$y = 4x^{-1} + 3x^{-2} + 2x^{-3}$$

$$\frac{dy}{dx} = 4(-1)x^{-1-1} + 3(-2)x^{-2-1} + 2(-3)x^{-3-1}$$ $$\frac{dy}{dx} = -4x^{-2} - 6x^{-3} - 6x^{-4}$$

$$\frac{dy}{dx} = -\frac{4}{x^2} - \frac{6}{x^3} - \frac{6}{x^4}$$

The final answer is $\boxed{-\frac{4}{x^2} - \frac{6}{x^3} - \frac{6}{x^4}}$.

4) Given $y = (-2x^2)^3$

$$y = (-2)^3 (x^2)^3$$ $$y = -8x^{2 \times 3}$$ $$y = -8x^6$$

$$\frac{dy}{dx} = -8(6)x^{6-1}$$ $$\frac{dy}{dx} = -48x^5$$

The final answer is $\boxed{-48x^5}$.

5) Given $y = (2x^2 - 3)(3x - 2)$

$$y = (2x^2)(3x) + (2x^2)(-2) + (-3)(3x) + (-3)(-2)$$ $$y = 6x^3 - 4x^2 - 9x + 6$$

$$\frac{dy}{dx} = 6(3)x^{3-1} - 4(2)x^{2-1} - 9(1)x^{1-1} + 0$$ $$\frac{dy}{dx} = 18x^2 - 8x - 9x^0$$ $$\frac{dy}{dx} = 18x^2 - 8x - 9$$

The final answer is $\boxed{18x^2 - 8x - 9}$.