Given: $(x^2 + y^2)^2 = xy$

Differentiate both sides with respect to $x$:

$\frac{d}{dx} (x^2 + y^2)^2 = \frac{d}{dx} (xy)$

Using the chain rule on the left side:

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

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

Expanding the left side:

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

Rearranging to solve for $\frac{dy}{dx}$:

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

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

Isolating $\frac{dy}{dx}$:

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