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Let's work through the given problems:

### (a) Differentiate with respect to \( x \)

#### (i) \( x^2 e^{3x + 2} \)
To differentiate this, we will use the **product rule** and **chain rule**.

The product rule states:
\[
\frac{d}{dx}(uv) = u'v + uv'
\]
where \( u = x^2 \) and \( v = e^{3x+2} \).

- \( u = x^2 \), so \( u' = 2x \).
- \( v = e^{3x+2} \), and by the chain rule, \( v' = e^{3x+2} \cdot 3 = 3e^{3x+2} \).

Using the product rule:
\[
\frac{d}{dx}(x^2 e^{3x+2}) = (2x)(e^{3x+2}) + (x^2)(3e^{3x+2}) = 2x e^{3x+2} + 3x^2 e^{3x+2}
\]
Thus, the derivative is:
\[
\boxed{2x e^{3x+2} + 3x^2 e^{3x+2}}
\]

#### (ii) \( \frac{\cos(2x^3)}{3x} \)
This requires the **quotient rule** and **chain rule**.

The quotient rule is:
\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v u' - u v'}{v^2}
\]
where \( u = \cos(2x^3) \) and \( v = 3x \).

- \( u = \cos(2x^3) \), and by the chain rule, \( u' = -\sin(2x^3) \cdot (6x^2) = -6x^2 \sin(2x^3) \).
- \( v = 3x \), so \( v' = 3 \).

Now applying the quotient rule:
\[
\frac{d}{dx} \left( \frac{\cos(2x^3)}{3x} \right) = \frac{(3x)(-6x^2 \sin(2x^3)) - (\cos(2x^3))(3)}{(3x)^2}
\]
Simplifying:
\[
= \frac{-18x^3 \sin(2x^3) - 3\cos(2x^3)}{9x^2}
\]
\[
= \frac{-6x^3 \sin(2x^3) - \cos(2x^3)}{3x^2}
\]
Thus, the derivative is:
\[
\boxed{\frac{-6x^3 \sin(2x^3) - \cos(2x^3)}{3x^2}}
\]

### (b) Given \( x = 4\sin(2y + 6) \), find \( \frac{dy}{dx} \) in terms of \( x \).

We will use **implicit differentiation**. Differentiate both sides with respect to \( x \):

\[
\frac{d}{dx}(x) = \frac{d}{dx}(4 \sin(2y + 6))
\]
The left side is:
\[
\frac{d}{dx}(x) = 1
\]

The right side requires the chain rule:
\[
\frac{d}{dx}(4\sin(2y+6)) = 4\cos(2y+6) \cdot \frac{d}{dx}(2y+6) = 8\cos(2y+6) \cdot \frac{dy}{dx}
\]

Thus, we have:
\[
1 = 8\cos(2y+6) \cdot \frac{dy}{dx}
\]

Solving for \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = \frac{1}{8\cos(2y+6)}
\]

To express \( \frac{dy}{dx} \) in terms of \( x \), recall that \( x = 4\sin(2y+6) \), so:
\[
\sin(2y+6) = \frac{x}{4}
\]

Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can solve for \( \cos(2y+6) \):
\[
\cos(2y+6) = \sqrt{1 - \sin^2(2y+6)} = \sqrt{1 - \left( \frac{x}{4} \right)^2} = \sqrt{\frac{16 - x^2}{16}} = \frac{\sqrt{16 - x^2}}{4}
\]

Substitute into the equation for \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} = \frac{1}{8 \cdot \frac{\sqrt{16 - x^2}}{4}} = \frac{1}{2\sqrt{16 - x^2}}
\]

Thus, the final result is:
\[
\boxed{\frac{dy}{dx} = \frac{1}{2\sqrt{16 - x^2}}}
\]

---

Would you like any additional details or explanations? Here are some follow-up questions you might find helpful:

1. How does the product rule work when differentiating products of functions?
2. What is the chain rule, and how does it apply in these problems?
3. Can you explain the quotient rule in further detail with an example?
4. How does implicit differentiation differ from regular differentiation?
5. What are the main trigonometric identities used in calculus?

**Tip**: When differentiating complex functions, always break them down step by step using the product, quotient, or chain rule to avoid mistakes!

Differentiate the following expressions: (a) (i) x^2 * e^(3x + 2), (ii) cos(2x^3) / (3x), (b) Given x = 4 * sin(2y + 6), find dy/dx in terms of x.

Learn how to differentiate complex expressions using the product, quotient, and chain rules. This problem also covers implicit differentiation with trigonometric functions. Perfect for students in Grades 11-12 or first-year college calculus.

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