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 \( F_{y}= \)

To evaluate \( F_y \) for the given function \( F(x, y) = 4 x^{3} + x^{2} y - 2 y^{3} \), we take the partial derivative of \( F \) with respect to \( y \): \[ F_y = \frac{\partial}{\partial y}(4 x^{3} + x^{2} y - 2 y^{3}) = 0 + x^{2} - 6 y^{2} = x^{2} - 6 y^{2}. \] Now that we have \( F_y = x^{2} - 6 y^{2} \), we can use this to find \( \frac{dy}{dx} \) using implicit differentiation. Next, applying implicit differentiation to \( F(x, y) = 0 \): \[ F_x \frac{dx}{dx} + F_y \frac{dy}{dx} = 0, \] we plug in the previously calculated derivatives: \[ 12 x^{2} + 2 x y + (x^{2} - 6 y^{2}) \frac{dy}{dx} = 0. \] Now simplify to find \( \frac{dy}{dx} \): \[ (x^{2} - 6 y^{2}) \frac{dy}{dx} = - (12 x^{2} + 2 x y), \] thus, \[ \frac{dy}{dx} = \frac{-(12 x^{2} + 2 x y)}{x^{2} - 6 y^{2}}. \]