Find y'' using implicit differentiation Pretty sure I messed up
$$
x^3 - 3xy + y^3 = 1
$$
$$
3x^2-3xy'-3y+3y^2y'=0
$$
$$
3y^2y'-3xy'=3y-3x^2
$$
$$
(3y^2-3x)y'=3y-3x^2
$$
$$
y' = \dfrac{y-x^2}{y^2-x}
$$
$$
y'' = \dfrac{(y^2-x)(y'-2x)-(y-x^2)(2yy'-1)}{(y^2-x)^2}
$$
$$
y'' = \dfrac{y^2y'-2xy^2-xy'+2x^2-2y^2y'+y+2x^2yy'-x^2}{(y^2-x)^2}
$$
$$
y''=\dfrac{x^2-2xy^2+y+(2x^2y-x-y^2)y'}{(y^2-x)^2}
$$
After subbing in $y'$, expanding and simplifying...
$$
y'' = \dfrac{x^2-2xy^2+y+3x^2y^2-2x^4y-xy+x^3-y^3}{(y^2-x)^3}
$$
The answer should be 
$$
y'' = -\dfrac{4xy}{(y^2-x)^3}
$$

As pointed out in the comments, the error occurred when plugging back in the value for $y'$.  The part that reads:
$$x^2-2xy^2+y$$
Should read:
$$(x^2-2xy^2+y)(y^2-x)$$
This multiplies out and helps simplify further.
