If x^3 + y^3 = 1, find an expression for y".?

By EXPLICIT DIFFERENTIATION : - [ Your L.H.S. sol. ]
. . .-----------------------------------------
x³+ y³ = 1
y³ = 1-x³
y = (1-x³)^(1/3) - - - - - - - (#)

(a)
i) If y ' is expressed in terms of x :
then y ' = [(1/3)(1-x³)^(-2/3)]*(-3x²) = -x² (1-x³)^(-2/3)

ii) If y ' is to be expressed in terms of x, y :
then y ' = -x² (1-x³)^(-2/3) = -x² [(1-x³)^(1/3)]⁻² = -x² y⁻² . . . . . . . (from #)

(b)
i) If y ' is expressed in terms of x :
then y '' = -x² d/dx[(1-x³)^(-2/3)] - (1-x³)^(-2/3) d/dx(x²)
= -x² (-2/3) * (1-x³)^(-5/3) - 2x * (1-x³)^(-2/3)
= (2/3)x (1-x³)^(-5/3) [x - 3(1-x³)]
= (2/3)x (1-x³)^(-5/3) * (-3+x+3x³)
= (2x/3)(-3+x+3x³) * (1-x³)^(-5/3)

ii) If y '' is to be expressed in terms of x, y :
then y '' = (2x/3)(-3+x+3x³) * [(1-x³)^(1/3)]⁻⁵ = (2x/3)(-3+x+3x³) * y⁻⁵ . . . . . . . (from #)
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By IMPLICIT DIFFERENTIATION : - [ Your R.H.S. sol. ]
. . .-----------------------------------------
i) x³+ y³ = 1
d/dx (x³ + y³) = d/dx (1)
d/dx (x³) + d/dx (y³) = 0
3x² + 3y² (dy/dx) = 0　
x² + y² y ' = 0 . . . . . . . . . . . [ ∵ dy/dx = y ' ]
∴ y ' = - x²/y²

但這方法是 Form 6 - U1 level,你可能未學！

ii) d/dx (y ') =d/dx(- x²/y²)
. . . . . . . . . . . . . . . . . .

圖中兩個也是錯誤的，左方第一次微分開始錯誤的，而右方的d/dy是不能像平時的做分數一樣
x^3+y^3=1
y^3=1-x^3
(3y^2)(dy/dx)=-3x^2
(y^2)(dy/dx)=-x^2
(2y)(d^2 y/dx^2)=-2x
dy^2 /dx^2=-x/y
y''=-x/y