
3x^3 + 4y^3 = 1
9x^2 + 12y^2 y' = 0
y' = 3x^2/4y^2
now, using the quotient rule,
y'' = (24xy^2 + 3x^2(8yy'))/ 16y^4
= (3xy^2 + 3x^2yy')/2y^4
= 3x (y  xy')/2y^3
= 3x(y  x*(3x^2/4y^2))/2y^3
= 3x(4y^3 + 3x^3)/8y^5
or, you can use implicit differentiation twice:
9x^2 + 12y^2 y' = 0
18x + 24y (y')^2 + 12y^2 y'' = 0
3x + 4y y'^2 + 2y^2 y'' = 0
2y^2 y'' = (3x + 4y (y')^2)
y'' = (3x + 4y (y')^2)/2y^2
= (3x + 4y * 9x^4/16y^4)/2y^2
= 3x(4y^3 + 3x^3)/8y^5posted by Steve
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