The equation provided is:

$4x^2y = -4$

We are tasked with finding $\frac{dy}{dx}$, which involves differentiating both sides of the equation with respect to $x$. We'll use implicit differentiation since $y$ is a function of $x$.

Step-by-step process:

1. Differentiate both sides of the equation:

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

2. Use the product rule for the left-hand side, since $4x^2y$ is a product of two functions: $4x^2$ and $y$.

   The product rule is: $\frac{d}{dx}(uv) = u'v + uv'$ Here, $u = 4x^2$ and $v = y$.

   Differentiating $u = 4x^2$: $\frac{du}{dx} = 8x$

   So, applying the product rule: $8xy + 4x^2\frac{dy}{dx}$

3. Differentiate the right-hand side:

   Since $-4$ is a constant, its derivative is 0: $\frac{d}{dx}(-4) = 0$

4. Set up the equation:

   After differentiating both sides, we get: $8xy + 4x^2\frac{dy}{dx} = 0$

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

   Subtract $8xy$ from both sides: $4x^2\frac{dy}{dx} = -8xy$

   Now divide both sides by $4x^2$ (assuming $x \neq 0$): $\frac{dy}{dx} = \frac{-8xy}{4x^2}$

   Simplifying the right-hand side: $\frac{dy}{dx} = \frac{-2y}{x}$

Thus, the derivative is:

$\frac{dy}{dx} = \frac{-2y}{x}$

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Would you like more details on any of the steps? Here are 5 related questions to consider:

1. How do you apply implicit differentiation to more complex equations?
2. Can you explain the product rule in more detail?
3. What happens if $x = 0$ in this equation?
4. How do you solve implicit differentiation problems with higher powers?
5. Can implicit differentiation be used with trigonometric functions?

Tip: Always simplify your equations as much as possible after differentiating, it helps avoid unnecessary complexity later on!