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 \( F_{y}=x^{2}-6 y^{2} \). \( \frac{d y}{d x}= \)

To find \(\frac{dy}{dx}\), we can use implicit differentiation based on the given function \(F(x, y) = 4x^{3} + x^{2}y - 2y^{3} = 0\). Using the implicit differentiation formula, we have: \[ \frac{df}{dx} = F_x + F_y \frac{dy}{dx} = 0 \] Substituting \(F_x\) and \(F_y\): \[ 12x^{2} + 2xy + (x^{2} - 6y^{2}) \frac{dy}{dx} = 0 \] Now, isolate \(\frac{dy}{dx}\): \[ (x^{2} - 6y^{2}) \frac{dy}{dx} = - (12x^{2} + 2xy) \] Solving for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = -\frac{12x^{2} + 2xy}{x^{2} - 6y^{2}} \] Thus, the final result is: \[ \frac{dy}{dx} = -\frac{12x^{2} + 2xy}{x^{2} - 6y^{2}} \]