Calcules
posted by Nurcan .
find y" by implicit differentiation.
4x^3 + 3y^3 = 4
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4x^3 + 3y^3 = 4
12x + 9yy' = 0
24x + 9(2yy')y' + 9y^2(y") = 0
y" = 6[4x3y(y')^2] answer. (it this right? )

Your 1st deriv should have been
12x^2 + 9y^2 y' = 0 > y' = 12x^2/(9y^2)
2nd derivative:
24x + 9y^2 (y'') + y' (18y) = 0
replacing y'
24x + 9y^2 y'' + (12x^2)/(9y^2) (18y) = 0
24x + 9y^2 y''  24x/y = 0
times y
24xy + 9y^3  24x = 0
y'' = (24x  24xy)/(9y^3) = (8x8xy)/(3y^3)
I don't think they wanted to see that y' in your answer.
btw, how did you ever get you first derivative ?