The equation is [tex]4x^3+5y^3=9[/tex].
Differentiating with respect to x, we have
[tex]12x^2+5(3y^2)(y')=0[/tex],
that is
[tex]12x^2+15y^2y'=0[/tex].
This means that [tex]\displaystyle{y'= -\frac{12x^2}{15y^2}=- \frac{4x^2}{5y^2} [/tex]
Differentiating again, using the quotient rule for the right hand side expression:
[tex]\displaystyle{y''=- \frac{4}{5}\cdot \frac{2xy^2-2yy'x^2}{y^4}= \frac{-8xy^2+8yy'x^2}{5y^4} [/tex]
