take first deritiivve
9x²+12y²(dy/dx)=0
minus 9x² both sides
12y²(dy/dx)=-9x²
divide both sides by 12y²
[tex] \frac{dy}{dx}= \frac{-9x^2}{12y^2} [/tex]
[tex] \frac{dy}{dx}= \frac{-3x^2}{4y^2} [/tex]
now take the deritive of right side
[tex] \frac{(-6x)(4y^2)-(8y)(\frac{dy}{dx})(-3x^2)}{(4y^2)^2} [/tex]
[tex] \frac{(-6x)(4y^2)-(8y)(\frac{-3x^2}{4y^2})(-3x^2)}{(4y^2)^2} [/tex]
[tex] \frac{-24xy^2-(\frac{18x^4}{y}}{16y^4} [/tex]
[tex] \frac{dy^2}{dx^2}= y''= \frac{-3x}{2}- \frac{9x^4}{8y^5} [/tex]
