Given the equation \( 2 x^{3}+x^{2} y-4 y^{3}=0 \), evaluate \( \frac{d y}{d x} \). Assume that the equation implicitly defines \( y \) as a differentiable function of \( x \). \( \frac{d y}{d x}=\square \)

To find \( \frac{dy}{dx} \) from the implicit equation \( 2x^3 + x^2y - 4y^3 = 0 \), we will use implicit differentiation. Differentiating both sides of the equation with respect to \( x \): \[ \frac{d}{dx}(2x^3) + \frac{d}{dx}(x^2y) - \frac{d}{dx}(4y^3) = 0 \] Using the product rule on \( x^2y \): \[ 6x^2 + (2xy + x^2\frac{dy}{dx}) - 12y^2\frac{dy}{dx} = 0 \] Now, we rearrange the equation to isolate \( \frac{dy}{dx} \): \[ 6x^2 + 2xy + x^2\frac{dy}{dx} - 12y^2\frac{dy}{dx} = 0 \] Combine the \( \frac{dy}{dx} \) terms: \[ 6x^2 + 2xy = -\left(x^2 - 12y^2\right) \frac{dy}{dx} \] Now, solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = -\frac{6x^2 + 2xy}{x^2 - 12y^2} \] Thus, the final result is: \[ \frac{dy}{dx} = -\frac{6x^2 + 2xy}{x^2 - 12y^2} \]