We first have to find the first derivative before we can find the second. First derivative with respect to x is [tex]9x^2+12y^2 \frac{dy}{dx}=0 [/tex] so [tex]12y^2 \frac{dy}{dx}=-9x^2 [/tex] and [tex] \frac{dy}{dx}=- \frac{9x^2}{12y^2}=- \frac{3x^2}{4y^2} [/tex]. That's the first derivative. Now for the second it's probably easiest to use the quotient rule, even though it's usually long and drawn out. That looks like this: [tex] \frac{dy}{dx}= \frac{4y^2(-6x)-[-3x^2(8y \frac{dy}{dx})] }{(4y^2)^2} [/tex] which simplifies a bit to [tex] \frac{dy}{dx}= \frac{-24xy^2+24x^2y( \frac{dy}{dx}) }{16y^4} [/tex]. We will multiply both sides by that denominator to get rid of it which leaves us with [tex]16y^4 \frac{dy}{dx}=-24xy^2+24x^2y \frac{dy}{dx} [/tex]. Get both dy/dx terms on the same side, and then factor it out. [tex] \frac{dy}{dx}(16y^4-24x^2y)=-24xy^2 [/tex]. Divide to isolate the dy/dx: [tex] \frac{d^2y}{dx^2}= \frac{-24xy^2}{16y^4-24x^2y} [/tex]. There's no simplifying you could do after that that would make any significant difference.
