To find the derivative of $y$ with respect to $x$ from the given relation

$(x^2 + y^2)^2 = ay^3,$

we'll use implicit differentiation. Here's the step-by-step process:

Step 1: Differentiate both sides of the equation with respect to $x$.

On the left-hand side, we need to apply the chain rule to $(x^2 + y^2)^2$. On the right-hand side, we differentiate $ay^3$ with respect to $x$.

Left-hand side: $(x^2 + y^2)^2$

Using the chain rule:

$\frac{d}{dx} \left( (x^2 + y^2)^2 \right) = 2(x^2 + y^2) \cdot \frac{d}{dx} (x^2 + y^2).$

Now, differentiate $x^2 + y^2$:

$\frac{d}{dx} (x^2 + y^2) = 2x + 2y \cdot \frac{dy}{dx}.$

Thus, the derivative of the left-hand side becomes:

$2(x^2 + y^2)(2x + 2y \cdot \frac{dy}{dx}).$

Right-hand side: $ay^3$

Now, differentiate $ay^3$ using the chain rule:

$\frac{d}{dx} (ay^3) = a \cdot 3y^2 \cdot \frac{dy}{dx}.$

Step 2: Set up the equation.

Now that we have differentiated both sides, the equation becomes:

$2(x^2 + y^2)(2x + 2y \cdot \frac{dy}{dx}) = 3ay^2 \cdot \frac{dy}{dx}.$

Step 3: Solve for $\frac{dy}{dx}$.

Distribute on the left-hand side:

$2(x^2 + y^2)(2x) + 2(x^2 + y^2)(2y \cdot \frac{dy}{dx}) = 3ay^2 \cdot \frac{dy}{dx}.$

Simplify the terms:

$4x(x^2 + y^2) + 4y(x^2 + y^2) \cdot \frac{dy}{dx} = 3ay^2 \cdot \frac{dy}{dx}.$

Now, collect all terms with $\frac{dy}{dx}$ on one side:

$4y(x^2 + y^2) \cdot \frac{dy}{dx} - 3ay^2 \cdot \frac{dy}{dx} = -4x(x^2 + y^2).$

Factor out $\frac{dy}{dx}$:

$\left( 4y(x^2 + y^2) - 3ay^2 \right) \cdot \frac{dy}{dx} = -4x(x^2 + y^2).$

Finally, solve for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \frac{-4x(x^2 + y^2)}{4y(x^2 + y^2) - 3ay^2}.$

This is the derivative of $y$ with respect to $x$, expressed implicitly.

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Would you like a more detailed explanation of any part of the solution? Let me know if you need any clarification or have further questions.

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Here are five related questions to explore:

1. How do you apply the chain rule in implicit differentiation?
2. What is the method for differentiating a product of two functions?
3. How do you differentiate terms involving both $x$ and $y$?
4. What is the significance of the derivative $\frac{dy}{dx}$ in terms of the slope of the curve?
5. How would you solve implicit equations that involve higher-order derivatives?

Tip: When dealing with implicit differentiation, always remember to differentiate terms involving $y$ as if they are functions of $x$, applying the chain rule for $y$ terms.