Use implicit differentiation to find $\frac{d y}{d x}$.
\[
\begin{array}{l}
y^{2}+3 x^{3}=9 y-7 x^{2} \\
\frac{d y}{d x}=
\end{array}
\]

Step 1 ：First, we differentiate both sides of the equation with respect to \(x\).

Step 2 ：Differentiating \(y^{2}\) with respect to \(x\) gives \(2y \frac{dy}{dx}\).

Step 3 ：Differentiating \(3x^{3}\) with respect to \(x\) gives \(9x^{2}\).

Step 4 ：Differentiating \(9y\) with respect to \(x\) gives \(9 \frac{dy}{dx}\).

Step 5 ：Differentiating \(-7x^{2}\) with respect to \(x\) gives \(-14x\).

Step 6 ：Putting it all together, we get \(2y \frac{dy}{dx} + 9x^{2} = 9 \frac{dy}{dx} - 14x\).

Step 7 ：We can rearrange this equation to isolate \(\frac{dy}{dx}\) on one side. This gives \(\frac{dy}{dx} = \frac{9x^{2} + 14x}{2y - 9}\).

Step 8 ：So, the derivative of \(y\) with respect to \(x\) is \(\boxed{\frac{9x^{2} + 14x}{2y - 9}}\).