Find y'' by implicit differentiation: x^3 + y^3 = 1

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Find y'' by implicit differentiation: x^3 + y^3 = 1

Mathematics
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ok now take the derivative again. Don't forget to use this statement in again for y'
can you help me precal please
http://www.wolframalpha.com/input/?i=find+the+derivative+of+y%3D%28-x%5E2%29%2Fy%5E2
Here is the final solution from wolfram alpha.
\[x^3 + y^3 = 1\] \[3x^2+3y^2y'=0\] \[y'=-\frac{x^2}{y^2}\]
\[y''=\frac{2xy^2-2x^2yy'}{y^4}\] by the quotient rule
replace \(y'\) by \(-\frac{x^2}{y^2}\) in the above, do some algebra and get the solution
got to that but i'm stuck anyway
\[y''=\frac{2xy^2-2x^2y\times \frac{x^2}{y^2}}{y^4}\]
multiply top and bottom by \(y^2)\] so clear the fraction then cancel some y's
typo there i lost the minus sign, should be \[y''=\frac{2xy^2+2x^2y\times \frac{x^2}{y^2}}{y^4}\]
\[y''=\frac{2xy^4+2x^4y}{y^6}\] \[y''=\frac{2xy^3+2x^4}{y^5}\] \[y''=\frac{2x(y^3+x^3)}{y^5}\] \[y''=\frac{2x}{y^5}\] since \(x^36y^3=1\)
rather since \(x^3+y^3=1\)

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