To solve the problem, we will use implicit differentiation to find the derivative of the given equation $x^2 + y^2 - y^2x^2 = 9$ and compute the slope at the point $(0, 3)$.

Step-by-Step Solution:

1. Original equation: $x^2 + y^2 - y^2x^2 = 9$

2. Differentiate implicitly with respect to $x$: Using the product rule and chain rule: $\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) - \frac{d}{dx}(y^2x^2) = \frac{d}{dx}(9)$ $2x + 2y\frac{dy}{dx} - \left[ (x^2)\frac{d}{dx}(y^2) + (y^2)\frac{d}{dx}(x^2) \right] = 0$

3. Simplify each term: $2x + 2y\frac{dy}{dx} - \left[ (x^2)(2y\frac{dy}{dx}) + (y^2)(2x) \right] = 0$ Expand: $2x + 2y\frac{dy}{dx} - \left[ 2x^2y\frac{dy}{dx} + 2xy^2 \right] = 0$

4. Group terms involving $\frac{dy}{dx}$: $2y\frac{dy}{dx} - 2x^2y\frac{dy}{dx} = -2x + 2xy^2$ Factor out $\frac{dy}{dx}$ on the left-hand side: $\frac{dy}{dx} (2y - 2x^2y) = -2x + 2xy^2$

5. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-2x + 2xy^2}{2y - 2x^2y}$

6. Evaluate at $(0, 3)$: Substitute $x = 0$ and $y = 3$: $\frac{dy}{dx} = \frac{-2(0) + 2(0)(3^2)}{2(3) - 2(0^2)(3)}$ Simplify: \frac{dy}{\frac{dy}{dx} = \frac{0}{6} = 0

Final Answer:

The slope of the graph of the given equation at the point $(0, 3)$ is:

$\boxed{0}$

Do you have any questions or want further clarification? Here are some follow-up questions you might consider:

1. How is implicit differentiation used when both $x$ and $y$ appear in the equation?
2. Can you explain how the product rule was applied to $y^2x^2$?
3. What happens if the point provided doesn't satisfy the original equation?
4. Can we verify the solution using a graph or alternative method?
5. How would the process differ if we were finding the second derivative?

Tip: When solving implicit differentiation problems, always organize your terms carefully, especially when grouping expressions involving $\frac{dy}{dx}$.