\( 1 \leftarrow \quad \begin{array}{l}\text { Given the equation } 4 x^{3}+x^{2} y-2 y^{3}=0 \text {, evaluate } \frac{d y}{d x} \text {. Assume that the equation implicitly defines } y \text { as a differentiable } 1 \\ \text { function of } x \text {. } \\ \text { If } F(x, y)=4 x^{3}+x^{2} y-2 y^{3} \text {, then } F x=\square .\end{array} \)

To find \(\frac{dy}{dx}\) from the given equation \(4x^3 + x^2y - 2y^3 = 0\), we can use implicit differentiation. First, we differentiate both sides of the equation with respect to \(x\): \[ \frac{d}{dx}(4x^3) + \frac{d}{dx}(x^2y) - \frac{d}{dx}(2y^3) = 0. \] This gives: \[ 12x^2 + (2xy + x^2\frac{dy}{dx}) - 6y^2\frac{dy}{dx} = 0. \] Rearranging, we find: \[ x^2\frac{dy}{dx} - 6y^2\frac{dy}{dx} = -12x^2 - 2xy. \] Factoring out \(\frac{dy}{dx}\): \[ \frac{dy}{dx}(x^2 - 6y^2) = -12x^2 - 2xy. \] Finally, solving for \(\frac{dy}{dx}\), we arrive at: \[ \frac{dy}{dx} = \frac{-12x^2 - 2xy}{x^2 - 6y^2}. \] Now for \(F_x\): To find \(F_x\), take the partial derivative of \(F(x, y)\): \[ F_x = \frac{\partial}{\partial x}(4x^3 + x^2y - 2y^3) = 12x^2 + 2xy. \]