Given the function:

$xy^2 + xy^2 + 4y^2 = 0$

Let's simplify it first.

1. Combine like terms: $2xy^2 + 4y^2 = 0$

2. Factor out $y^2$: $y^2(2x + 4) = 0$

This equation implies two possibilities:

1. $y^2 = 0$, which gives $y = 0$.
2. $2x + 4 = 0$, which simplifies to $x = -2$.

Since we want to find $\frac{dy}{dx}$, we need to proceed with implicit differentiation on the original equation to capture the derivative relationship between $x$ and $y$.

Differentiating Implicitly

Starting from: $2xy^2 + 4y^2 = 0$

Differentiate each term with respect to $x$:

1. For $2xy^2$: $\frac{d}{dx}(2xy^2) = 2 \left( y^2 + x \cdot 2y \frac{dy}{dx} \right) = 2y^2 + 4xy \frac{dy}{dx}$

2. For $4y^2$: $\frac{d}{dx}(4y^2) = 4 \cdot 2y \frac{dy}{dx} = 8y \frac{dy}{dx}$

Thus, the differentiated equation is: $2y^2 + 4xy \frac{dy}{dx} + 8y \frac{dy}{dx} = 0$

Solving for $\frac{dy}{dx}$

Combine terms involving $\frac{dy}{dx}$: $2y^2 + (4xy + 8y) \frac{dy}{dx} = 0$

Move $2y^2$ to the other side: $(4xy + 8y) \frac{dy}{dx} = -2y^2$

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

Simplify by factoring out $y$ from both the numerator and denominator: $\frac{dy}{dx} = \frac{-2y}{4x + 8} = \frac{-y}{2x + 4}$

Final Answer

$\frac{dy}{dx} = \frac{-y}{2x + 4}$

Would you like further details or have any questions about the steps?

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Here are some related questions to explore:

1. How would the solution change if the equation was $xy^2 + 3xy^2 + 4y^2 = 0$?
2. What is the interpretation of $\frac{dy}{dx}$ if $x = -2$?
3. How would you find $\frac{d^2y}{dx^2}$ for this function?
4. What happens to $\frac{dy}{dx}$ if $y$ approaches zero?
5. How does implicit differentiation help in finding derivatives of non-standard functions?

Tip: When differentiating implicitly, keep terms involving $\frac{dy}{dx}$ on one side to simplify solving.