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

Step 1 ：Given the equation $F(x, y)=3 x^{3}+2 x^{2} y-4 y^{3}=0$, we are asked to find $\frac{d y}{d x}$.

Step 2 ：First, we need to find the partial derivatives $F_{x}$ and $F_{y}$.

Step 3 ：Taking the partial derivative of $F$ with respect to $x$, we get $F_{x}=9 x^{2}+4 x y$.

Step 4 ：Taking the partial derivative of $F$ with respect to $y$, we get $F_{y}=2 x^{2}-12 y^{2}$.

Step 5 ：According to the implicit function theorem, $\frac{d y}{d x}=-\frac{F_{x}}{F_{y}}$.

Step 6 ：Substituting $F_{x}$ and $F_{y}$ into the formula, we get $\frac{d y}{d x}=-\frac{9 x^{2}+4 x y}{2 x^{2}-12 y^{2}}$.

Step 7 ：Thus, the derivative of $y$ with respect to $x$ is $\boxed{-\frac{9 x^{2}+4 x y}{2 x^{2}-12 y^{2}}}$.