To find the second derivative, we need to differentiate the equation:

`x^3+y^3=125` differentiate - call this equation (1)

`3x^2+3y^2{dy}/{dx}=0` divide by 3

`x^2+y^2{dy}/{dx}=0` differentiate again, and use product rule on second term - call this equation (2)

`2x+2y{dy}/{dx}{dy}/{dx}+y^2{d^2y}/{dx^2}=0` simplify slightly

`2x+2y({dy}/{dx})^2+y^2{d^2y}/{dx^2}=0` equation (3)

Now from (1), we see that...

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To find the second derivative, we need to differentiate the equation:

`x^3+y^3=125` differentiate - call this equation (1)

`3x^2+3y^2{dy}/{dx}=0` divide by 3

`x^2+y^2{dy}/{dx}=0` differentiate again, and use product rule on second term - call this equation (2)

`2x+2y{dy}/{dx}{dy}/{dx}+y^2{d^2y}/{dx^2}=0` simplify slightly

`2x+2y({dy}/{dx})^2+y^2{d^2y}/{dx^2}=0` equation (3)

Now from (1), we see that when x=0, y=5

And from (2), when x=0 and y=5, then `{dy}/{dx}|_{x=0}=0` .

Now sub in the results to (3) to get:

`{d^2y}/{dx^2}|_{x=0}=0`

**Using implicit differentiation, the result is `{d^2y}/{dx^2}|_{x=0}=0` .**