The problem asks you to find the second derivative, $\frac{d^2y}{dx^2}$, implicitly using the equation:

$-4x^2 + 3y^2 + 4y = 0$

at the point $(1, -2)$.

Steps:

1. Differentiate the given equation implicitly with respect to $x$:

   $\frac{d}{dx} \left( -4x^2 + 3y^2 + 4y \right) = 0$

   Apply the chain rule to differentiate $y^2$ and $y$ terms:

   $-8x + 6y \frac{dy}{dx} + 4 \frac{dy}{dx} = 0$

   Combine terms:

   $-8x + (6y + 4) \frac{dy}{dx} = 0$

2. Solve for $\frac{dy}{dx}$:

   $\frac{dy}{dx} = \frac{8x}{6y + 4}$

   At the point $(1, -2)$:

   $\frac{dy}{dx} = \frac{8(1)}{6(-2) + 4} = \frac{8}{-12 + 4} = \frac{8}{-8} = -1$

3. Differentiate implicitly again to find the second derivative $\frac{d^2y}{dx^2}$:

   Differentiate $\frac{dy}{dx} = \frac{8x}{6y + 4}$ again with respect to $x$, applying the quotient rule:

   $\frac{d}{dx} \left( \frac{8x}{6y + 4} \right)$

   Apply the quotient rule $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$, where $u = 8x$ and $v = 6y + 4$:

   $\frac{d^2y}{dx^2} = \frac{(6y + 4)(8) - 8x(6 \frac{dy}{dx})}{(6y + 4)^2}$

   Plug in $x = 1$, $y = -2$, and $\frac{dy}{dx} = -1$:

   $\frac{d^2y}{dx^2} = \frac{(-12 + 4)(8) - 8(1)(6)(-1)}{(-12 + 4)^2}$

   Simplify the numerator:

   $(-8)(8) - 8(6)(-1) = -64 + 48 = -16$

   And the denominator:

   $(-8)^2 = 64$

   Thus:

   $\frac{d^2y}{dx^2} = \frac{-16}{64} = -\frac{1}{4}$

Final Answer:

$\frac{d^2y}{dx^2} = -\frac{1}{4}$

Would you like any additional details or explanations?

Here are 5 related questions you might find helpful:

1. How do you apply the quotient rule in implicit differentiation?
2. Why do we need to differentiate twice to find the second derivative?
3. Can implicit differentiation always be used to find higher-order derivatives?
4. How does the chain rule affect the differentiation of $y^2$?
5. What would change if the point was $(1, 2)$ instead of $(1, -2)$?

Tip: Always check your first derivative before proceeding to the second derivative in implicit differentiation!