Given the equation:

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

We need to find $\frac{dy}{dx}$ using implicit differentiation.

Step-by-step solution:

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

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

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

   For the term $(x^2 - y^2)^3$, we first differentiate the outer function, treating $x^2 - y^2$ as a single term:

   $3(x^2 - y^2)^2 \cdot \frac{d}{dx} (x^2 - y^2)$

3. Differentiate $x^2 - y^2$ with respect to $x$:

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

   (The term $2y \frac{dy}{dx}$ is due to the chain rule, as $y$ is a function of $x$.)

4. Substitute the differentiated expression:

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

5. Simplify the equation:

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

   Distribute the $3$:

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

6. Divide both sides by 6:

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

7. Expand the left-hand side:

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

8. Isolate the $\frac{dy}{dx}$ term:

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

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

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

Thus, the derivative of $y$ with respect to $x$ is:

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

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Would you like further details or have any questions about this solution?

Here are some related questions for practice:

1. How would you apply implicit differentiation to more complex equations?
2. What is the significance of the chain rule in implicit differentiation?
3. Can you solve for $\frac{dy}{dx}$ when the equation is more complicated, like involving trigonometric functions?
4. How does the method differ for an equation with multiple variables on both sides?
5. What is the relationship between implicit and explicit differentiation?

Tip: Always pay careful attention to terms involving $y$ during differentiation, as you must treat $y$ as a function of $x$ and apply the chain rule.