To solve this problem, we will proceed with the following steps for parts (a) and (b) of the question:

Given Equation:

$xy + 2x + 3x^2 = 9$

Part (a): Implicit Differentiation

1. Differentiate both sides of the equation with respect to $x$. Recall that $y$ is a function of $x$, so when differentiating $y$, we must apply the chain rule and include $y'$.

   Differentiating term by term: $\frac{d}{dx}(xy) + \frac{d}{dx}(2x) + \frac{d}{dx}(3x^2) = \frac{d}{dx}(9)$

2. Apply the product rule to the term $xy$: $\frac{d}{dx}(xy) = x \cdot \frac{dy}{dx} + y \cdot \frac{dx}{dx} = x y' + y$

3. Differentiate the remaining terms: $\frac{d}{dx}(2x) = 2 \quad \text{and} \quad \frac{d}{dx}(3x^2) = 6x$

4. Substitute these derivatives back into the differentiated equation: $x y' + y + 2 + 6x = 0$

5. Solve for $y'$: $x y' = -y - 2 - 6x$ $y' = \frac{-y - 2 - 6x}{x}$

So, the answer for part (a) is: $y' = \frac{-y - 2 - 6x}{x}$

Part (b): Solve for $y$ and Differentiate

1. Rearrange the original equation to solve for $y$: $xy + 2x + 3x^2 = 9$ $xy = 9 - 2x - 3x^2$ $y = \frac{9 - 2x - 3x^2}{x}$

2. Differentiate $y$ with respect to $x$: $y' = \frac{d}{dx} \left( \frac{9 - 2x - 3x^2}{x} \right)$

3. Use the quotient rule, where $u = 9 - 2x - 3x^2$ and $v = x$: $y' = \frac{v \cdot u' - u \cdot v'}{v^2}$

4. Compute $u'$ and $v'$: $u' = -2 - 6x \quad \text{and} \quad v' = 1$

5. Substitute these into the quotient rule: $y' = \frac{x(-2 - 6x) - (9 - 2x - 3x^2)(1)}{x^2}$

6. Simplify the expression: $y' = \frac{-2x - 6x^2 - 9 + 2x + 3x^2}{x^2}$ $y' = \frac{-9 - 3x^2}{x^2}$

Both parts should yield consistent results in terms of structure.

Would you like further details on any steps?

Here are some related questions:

1. How do we use implicit differentiation in other functions?
2. What does consistency between both methods indicate?
3. Why is the quotient rule useful in finding derivatives?
4. How does implicit differentiation handle terms with multiple variables?
5. Can we differentiate functions directly after simplifying them?

Tip: Consistency between two differentiation methods provides confidence in accuracy!