\begin{tabular}{c} Find \( \frac{d^{2} y}{d x^{2}} \) in terms of \( x \) and \( y \). \\ \( \qquad x^{2} y^{2}-12 x=9 \) \\ \( \frac{d^{2} y}{d x^{2}}=\square \) \\ \hline\end{tabular}

To find \(\frac{d^{2} y}{d x^{2}}\) in terms of \(x\) and \(y\), we begin by differentiating the given equation \(x^{2} y^{2} - 12x = 9\) with respect to \(x\). 1. **Differentiate Once:** Starting with: \[ x^{2} y^{2} - 12x - 9 = 0 \] We apply implicit differentiation: \[ 2x y^{2} + x^{2} (2y \frac{dy}{dx}) - 12 = 0 \] Simplifying gives: \[ 2xy^{2} + 2x^{2}y \frac{dy}{dx} - 12 = 0 \] Rearranging to solve for \(\frac{dy}{dx}\): \[ 2x^{2} y \frac{dy}{dx} = 12 - 2xy^{2} \] Thus, \[ \frac{dy}{dx} = \frac{12 - 2xy^{2}}{2x^{2} y} = \frac{6 - xy^{2}}{x^{2} y} \] 2. **Differentiate Again:** Now, we differentiate \(\frac{dy}{dx}\): \[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{6 - xy^{2}}{x^{2} y} \right) \] Using the quotient rule: \[ \frac{d^2y}{dx^2} = \frac{(x^{2}y)(-y^{2} - 2xy\frac{dy}{dx}) - (6 - xy^{2})(2xy + x^{2}\frac{dy}{dx})}{(x^{2}y)^{2}} \] Now simplify and substitute \(\frac{dy}{dx}\) into this equation, we can find \(\frac{d^{2}y}{dx^{2}}\) in terms of \(x\) and \(y\). 3. **Final Expression:** The second derivative simplifies down into a complex form involving \(x\), \(y\), and \(\frac{dy}{dx}\). The goal is to express it cleanly with proper substitutions and algebra manipulations. By this method, you will obtain \(\frac{d^{2}y}{dx^{2}}\) expressed accurately as you delve deeper into the calculation.