Please someone please find the second differential for this equation
(x^{3}) + (y^{3}) = 1
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well for te first differential i got
3x^{2} + 3y^{2}(dy/dx)=0
dy/dx = (-x^{2} )/(y^{2})
Then seoncd differntial using quotient rule
[(y^{2})(-2x)-(-x^{2} )(2y(dy/dx))]/(y^{4})
then finally the second differntial came down to for me
[-2xy^{3} + 2x^{4} ]/y^{5}
Please help