#x^(3)+y^(3)=1#
We use implicit differentiation:
#D[x^(3)+y^(3)]=D[1]#
So:
#3x^2+3y^2y'=0#
#y'=-(cancel(3)x^2)/(cancel(3)y^2)=-(x^2)/(y^(2)# #color(red)((1))#
So now we have to differentiate again using the quotient rule. A useful tip is to start and finish with the function on the bottom #rArr#
#y''=(y^(2)(-2x)-(-x^(2)).2y.y')/(y^(4))# #color(red)((2))#
We already know from #color(red)((1))# that :
#y'=-x^(2)/y^2#
So we can substitute that expression for #y'# into #color(red)((2))# #rArr#
#y''=(y^(2)(-2x)-(-x^(2)).2y.((-x^(2))/(y^(2))))/(y^(4)#
#y''=[-2xy^2-(2x^4)/(y)]/[y^4]#
Multiplying top and bottom by #y# #rArr#
#y''=([-2xy^3-2x^4])/(y^5)#
#y''=(-2x[y^(3)+x^(3)])/(y^5)# #color(red)((3))#
Since we know that:
#x^3+y^3=1#
We can substitute that value of #1# into #color(red)((3))rArr#
#y''=(-2x)/(y^5)#
