Given the equation \( 4 x^{3}+x^{2} y-2 y^{3}=0 \), evaluate \( \frac{d y}{d x} \). Assume that the equation implicitly defines \( y \) as a differentiable function of \( x \). If \( F(x, y)=4 x^{3}+x^{2} y-2 y^{3} \), then \( F_{x}=12 x^{2}+2 x y \). If \( F(x, y)=4 x^{3}+x^{2} y-2 y^{3} \), then \( F_{y}=\square \)

To find \( F_y \) for the function \( F(x, y) = 4x^3 + x^2 y - 2y^3 \), we will differentiate \( F \) with respect to \( y \): \[ F_y = \frac{\partial}{\partial y}(4x^3 + x^2 y - 2y^3) = x^2 - 6y^2. \] So, \( F_y = x^2 - 6y^2 \). Now, to evaluate \( \frac{dy}{dx} \) using the implicit differentiation formula, we have: \[ \frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{12x^2 + 2xy}{x^2 - 6y^2}. \] This expression gives us \( \frac{dy}{dx} \) in terms of \( x \) and \( y \).