1. **State the problem:** We need to find the second derivative $\frac{d^2y}{dx^2}$ implicitly from the equation $$x^2y - 4 = 9x + y.$$\n\n2. **Rewrite the equation:** $$x^2y - 4 = 9x + y.$$\n\n3. **Differentiate both sides with respect to $x$ implicitly:**\nUsing the product rule on $x^2y$, we get:\n$$\frac{d}{dx}(x^2y) = 2xy + x^2\frac{dy}{dx}.$$\nDifferentiating the right side:\n$$\frac{d}{dx}(9x + y) = 9 + \frac{dy}{dx}.$$\nSo the differentiated equation is:\n$$2xy + x^2\frac{dy}{dx} = 9 + \frac{dy}{dx}.$$\n\n4. **Group terms with $\frac{dy}{dx}$ on one side:**\n$$x^2\frac{dy}{dx} - \frac{dy}{dx} = 9 - 2xy.$$\n\n5. **Factor out $\frac{dy}{dx}$:**\n$$\frac{dy}{dx}(x^2 - 1) = 9 - 2xy.$$\n\n6. **Solve for $\frac{dy}{dx}$:**\n$$\frac{dy}{dx} = \frac{9 - 2xy}{x^2 - 1}.$$\n\n7. **Differentiate $\frac{dy}{dx}$ again to find $\frac{d^2y}{dx^2}$:**\nLet $$u = 9 - 2xy, \quad v = x^2 - 1.$$\nThen $$\frac{dy}{dx} = \frac{u}{v}.$$\nUsing the quotient rule:\n$$\frac{d^2y}{dx^2} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}.$$\n\n8. **Calculate $\frac{du}{dx}$:**\n$$\frac{du}{dx} = -2\left(y + x\frac{dy}{dx}\right) = -2y - 2x\frac{dy}{dx}.$$\n\n9. **Calculate $\frac{dv}{dx}$:**\n$$\frac{dv}{dx} = 2x.$$\n\n10. **Substitute into the quotient rule:**\n$$\frac{d^2y}{dx^2} = \frac{(x^2 - 1)(-2y - 2x\frac{dy}{dx}) - (9 - 2xy)(2x)}{(x^2 - 1)^2}.$$\n\n11. **Final answer:**\n$$\boxed{\frac{d^2y}{dx^2} = \frac{(x^2 - 1)(-2y - 2x\frac{dy}{dx}) - 2x(9 - 2xy)}{(x^2 - 1)^2}}$$\nwhere $$\frac{dy}{dx} = \frac{9 - 2xy}{x^2 - 1}.$$