bk2 p50 ex17c q20

given

$\displaystyle x^3 - 3xy +y^3 =4$

find

$\displaystyle \frac{d^2 y }{dx^2}$

my working:

i use y' for dy/dx

$\displaystyle y' = \frac {x^2-y}{x-y^2}$

$\displaystyle y''= \frac {(x-y^2)(2x-y')-(x^2-y)(1-2y y')}{(x-y^2)^2}$

$\displaystyle y''= \frac {(-x+2x^2-y^2)y'+x^2-2xy+y}{(x-y^2)^2}$

stuck here. thanks

p.s. why does that 'dash' -- y' look so bad in latex?