All you need to do is differentiate twice.
[tex] \frac{d}{dx} (4x + 3y) = \frac{d}{dx} \sin(y) \\ 4dx + 3dy = \cos(y)dy \\ (\cos(y) - 3)dy = 4dx \\ \frac{dy}{dx} = \frac{4}{ \cos(y) - 3} \\ \frac{ {d}^{2}y}{d{x}^{2}} = 4 {( \cos(y) - 3)}^{ - 1} \\ = - 4 {(\cos(y) - 3)}^{ - 2} \\ = - 4 {(\cos(y) - 3)}^{ - 2} ( - \sin(y)) \\ = \frac{4 \sin(y) }{ {( \cos(y) - 3)}^{2} } [/tex]
I first used implicit differentiation, then used the chain rule to differentiate again.
