1. **State the problem:** We need to find the first derivative $\frac{dy}{dx}$ and the second derivative $\frac{d^2y}{dx^2}$ implicitly from the equation $$3xy + \sin(x) = 9.$$\n\n2. **Recall implicit differentiation:** When $y$ is a function of $x$, differentiate both sides with respect to $x$, remembering to use the product rule for $3xy$ and chain rule for $y$.\n\n3. **Differentiate both sides:**\n$$\frac{d}{dx}(3xy) + \frac{d}{dx}(\sin(x)) = \frac{d}{dx}(9)$$\nUsing product rule on $3xy$: $$3\left(x\frac{dy}{dx} + y\right) + \cos(x) = 0.$$\n\n4. **Solve for $\frac{dy}{dx}$:**\n$$3x\frac{dy}{dx} + 3y + \cos(x) = 0$$\n$$3x\frac{dy}{dx} = -3y - \cos(x)$$\n$$\frac{dy}{dx} = \frac{-3y - \cos(x)}{3x}$$\nShow cancellation:\n$$\frac{\cancel{3}x\frac{dy}{dx}}{\cancel{3}x} = \frac{-3y - \cos(x)}{3x}$$\n\n5. **Find the second derivative $\frac{d^2y}{dx^2}$:** Differentiate $\frac{dy}{dx}$ implicitly:\n$$\frac{dy}{dx} = \frac{-3y - \cos(x)}{3x}$$\nRewrite as:\n$$y' = \frac{-3y - \cos(x)}{3x}$$\nUse quotient rule: If $y' = \frac{u}{v}$, then $$y'' = \frac{v u' - u v'}{v^2}.$$\nHere, $u = -3y - \cos(x)$ and $v = 3x$.\n\n6. **Compute derivatives:**\n$$u' = -3\frac{dy}{dx} + \sin(x)$$\n$$v' = 3$$\n\n7. **Apply quotient rule:**\n$$y'' = \frac{3x(-3y' + \sin(x)) - (-3y - \cos(x))3}{(3x)^2}$$\nSimplify numerator:\n$$= \frac{3x(-3y') + 3x\sin(x) + 9y + 3\cos(x)}{9x^2}$$\n\n8. **Substitute $y' = \frac{-3y - \cos(x)}{3x}$ into numerator:**\n$$3x(-3y') = 3x\left(-3 \cdot \frac{-3y - \cos(x)}{3x}\right) = 3x \cdot \frac{9y + 3\cos(x)}{3x} = 9y + 3\cos(x)$$\n\n9. **Numerator becomes:**\n$$9y + 3\cos(x) + 3x\sin(x) + 9y + 3\cos(x) = 18y + 6\cos(x) + 3x\sin(x)$$\n\n10. **Final expression for $y''$:**\n$$y'' = \frac{18y + 6\cos(x) + 3x\sin(x)}{9x^2} = \frac{2(9y + 3\cos(x)) + 3x\sin(x)}{9x^2}$$\nSimplify by dividing numerator and denominator by 3:\n$$y'' = \frac{6y + 2\cos(x) + x\sin(x)}{3x^2}.$$\n\n**Final answers:**\n$$\frac{dy}{dx} = \frac{-3y - \cos(x)}{3x}$$\n$$\frac{d^2y}{dx^2} = \frac{6y + 2\cos(x) + x\sin(x)}{3x^2}$$