To find the derivative of $y$ with respect to $x$ from the equation:

$(x^2 - y^2)^3 = 3a^4x^2$

follow these steps:

Solution Process

1. Differentiate both sides with respect to $x$:

   The left-hand side involves a composite function, so we'll use the chain rule.

   [ \frac{d}{dx}\left[ (x^2 - y^2)^3 \right] = \frac{d}{dx}\left[ 3a^4x^2 \right] ]

2. Apply the chain rule to the left-hand side:

   For the expression $(x^2 - y^2)^3$, let $u = x^2 - y^2$, so we differentiate:

   $\frac{d}{dx} \left[ u^3 \right] = 3u^2 \cdot \frac{du}{dx}$

   Now, calculate $\frac{du}{dx}$:

   $u = x^2 - y^2 \quad \Rightarrow \quad \frac{du}{dx} = 2x - 2y \frac{dy}{dx}$

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

   $3(x^2 - y^2)^2 \cdot \left( 2x - 2y \frac{dy}{dx} \right)$

3. Differentiate the right-hand side:

   $\frac{d}{dx}\left[ 3a^4x^2 \right] = 6a^4x$

4. Equating both sides:

   Now we have:

   $3(x^2 - y^2)^2 \cdot \left( 2x - 2y \frac{dy}{dx} \right) = 6a^4x$

5. Simplify and solve for $\frac{dy}{dx}$:

   $(x^2 - y^2)^2 \cdot (2x - 2y \frac{dy}{dx}) = 2a^4x$

   Expand the left-hand side:

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

6. Group the terms involving $\frac{dy}{dx}$:

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

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

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

   Simplifying further:

   $\frac{dy}{dx} = \frac{x\left[(x^2 - y^2)^2 - a^4\right]}{y(x^2 - y^2)^2}$

Final Answer:

$\frac{dy}{dx} = \frac{x\left[(x^2 - y^2)^2 - a^4\right]}{y(x^2 - y^2)^2}$

---

Would you like further clarification or additional steps? Here are some follow-up questions for deeper understanding:

1. How would you find the second derivative of $y$ with respect to $x$?
2. What happens if $y = 0$ in the given equation?
3. How does implicit differentiation differ from explicit differentiation?
4. What would the derivative be if the right-hand side had a different function of $x$, such as $\sin(x)$?
5. How can you apply this method to other implicit functions involving both $x$ and $y$?

Tip: When differentiating implicitly, remember that every time you differentiate a $y$-term, multiply by $\frac{dy}{dx}$ to account for the dependence of $y$ on $x$.