Show that y''=-(3y^2)/(y^7) if x^4-y^4=1.?

Sol
x^4－y^4=1
4x^3－4y^3y’=0
x^3－y^3y’=0
3x^2－y^3y”－3y^2(y’)^2=0
y^3y”=3x^2－3y^2(y’)^2
y^3y”=3x^2－3y^2(x^3/y^3)^2
y^3y”=3x^2－3y^2*(x^6/y^6)
y^7y”=3x^2y^4－3x^4
(y^7/x^2)y”=3y^4－3x^4
(y^7/x^2)y”=－3
y"=－3x^2/y^7

唔好意思，我計出的是
y" = -(3x^2)/(y^7)