You need to write the function in terms of x, hence, you need to isolate `y^3` to the left side such that:

`-y^3 = 3 - x^3 => y^3 = x^3 - 3 => y = root(3)(x^3 - 3)`

You need to convert the cube root into a power such that:

`y = (x^3 - 3)^(1/3)`

You need to differentiate the function with respect to x, using the chain rule, such that:

`(dy)/(dx) = (1/3)(x^3 - 3)^(1/3 - 1)(3x^2)`

`(dy)/(dx) = (1/3)(x^3 - 3)^(-2/3)(3x^2)`

`(dy)/(dx) = (x^2)(x^3 - 3)^(-2/3)`

You need to differentiate again to evaluate `(d^2y)/(dx^2)` such that:

`(d^2y)/(dx^2)= (x^2)'((x^3 - 3)^(-2/3)) + (x^2)((x^3 - 3)^(-2/3))'`

`(d^2y)/(dx^2) = 2x((x^3 - 3)^(-2/3)) + (x^2)((x^3 - 3)^(-2/3 - 1))*(3x^2)`

`(d^2y)/(dx^2) = 2x((x^3 - 3)^(-2/3)) + 3x^4((x^3 - 3)^(-5/3))`

You should factor out `x((x^3 - 3)^(-5/3))` such that:

`(d^2y)/(dx^2) = (x((x^3 - 3)^(-5/3)))(2((x^3 - 3)^(-2/3+5/3)) + 3x^3)`

`(d^2y)/(dx^2) = (x((x^3 - 3)^(-5/3)))(2(x^3 - 3 + 3x^3))`

`(d^2y)/(dx^2) = ((x^3 - 3)^(-5/3)))(8x^4 - 6x)`

**Hence, evaluating the second derivative using implicit differentiation yields `(d^2y)/(dx^2) = ((x^3 - 3)^(-5/3)))(8x^4 - 6x).` **

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