The equation provided is $x^2 y^2 - 18x = 4$, and we need to find $\frac{d^2 y}{dx^2}$ in terms of $x$ and $y$.

Step-by-step solution:

1. Implicit Differentiation (First derivative): Differentiate both sides of $x^2 y^2 - 18x = 4$ with respect to $x$, using the product rule for the term $x^2 y^2$.

   $\frac{d}{dx}\left( x^2 y^2 \right) - \frac{d}{dx}(18x) = \frac{d}{dx}(4)$

   Using the product rule on $x^2 y^2$:

   $\frac{d}{dx}(x^2 y^2) = 2x y^2 + x^2 \cdot 2y \frac{dy}{dx}$

   So, we have:

   $2x y^2 + 2x^2 y \frac{dy}{dx} - 18 = 0$

   Now, solve for $\frac{dy}{dx}$:

   $2x^2 y \frac{dy}{dx} = 18 - 2x y^2$

   $\frac{dy}{dx} = \frac{18 - 2x y^2}{2x^2 y}$

2. Second Derivative (Finding $\frac{d^2y}{dx^2}$): Now differentiate $\frac{dy}{dx}$ again with respect to $x$ to find $\frac{d^2 y}{dx^2}$.

   Using the quotient rule:

   $\frac{d}{dx} \left( \frac{18 - 2x y^2}{2x^2 y} \right)$

   We'll apply the quotient rule here, which states:

   $\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$

   Let $u = 18 - 2x y^2$ and $v = 2x^2 y$.

   • $\frac{du}{dx} = -2y^2 - 2x \cdot 2y \frac{dy}{dx} = -2y^2 - 4xy \frac{dy}{dx}$
   • $\frac{dv}{dx} = 4xy + 2x^2 \frac{dy}{dx}$

Now I'll calculate and simplify these terms to provide the final result.The second derivative $\frac{d^2y}{dx^2}$ is:

$\frac{d^2 y}{dx^2} = \frac{x y^2 - 18}{x^3 y}$

This is the result in terms of $x$ and $y$.

Would you like more details on the steps, or do you have any questions?

Here are 5 related questions to explore further:

1. How does the product rule apply in implicit differentiation?
2. Can implicit differentiation be used to find higher-order derivatives?
3. How is the quotient rule applied when differentiating fractions?
4. What are the practical applications of second derivatives in physics or economics?
5. How would the result change if the original equation involved different powers of $x$ or $y$?

Tip: Implicit differentiation is a powerful tool when dealing with equations where $y$ is a function of $x$ but isn't isolated.