Calculus: Triple derivative. Where am I going wrong?

Question

$$x^2+xy+y^3=1$$Find $y'''$ at $x = 1$

My solution:

$$y\Big|_{x = 1} = 0 \\ 2x + y + xy' + 3y^2y' = 0\\ 2 + y' + y' + xy'' + y'(6yy') + 3y^2y'' = 0 \\ 2 + 2y' + xy'' + 2y'6y + 3y^2y'' = 0 \\ y''\Big|_{x = 1} = \frac{-2-2y'-2y'6y}{3y^2 + x}\Big|_{x=1} = \frac{-2 -2(-2)}{3(0) + 1} = 2 \\ 2y'' + y'' + xy''' + 6y2y'' + 2y'6y' + y''6y' + 3y^2y''' = 0 \\ y''' = \frac{-3y'' -6y2y'' - 12y' - 6yy'y''}{x+3y^2} \\ y'''\Big|_{x = 1} = \frac{-3(2) - 12(-2)}{1} = \boxed{-30}$$

Answer 1

$y\Big|_{x = 1} = 0$ is so trivial. So you have missed 3rd to 4th line, simplified $y'6yy'$ to $2y'6y$ not $6yy'^2$ is wrong, but it seems, it does not affects on answer... but verification is first, always.

Then...

$$ 0=\frac d{dx}1=\frac d{dx}(x^2+xy+y^3)=2x+y+xy'+3y^2y' \\ 0=2+y\Big|_{x = 1}+y'\Big|_{x = 1}+3\left(y\Big|_{x = 1}\right)^2y'\Big|_{x = 1}=2+y'\Big|_{x = 1} \\ y'\Big|_{x = 1}=-2 \\ 0=\frac d{dx}0=\frac d{dx}(2x+y+xy'+3y^2y')=2+2y'+xy''+6y(y')^2+3y^2y'' \\ 0=2+2y'\Big|_{x = 1}+y''\Big|_{x = 1}+6y\Big|_{x = 1}\left(y'\Big|_{x = 1}\right)^2+3\left(y\Big|_{x = 1}\right)^2y''\Big|_{x = 1} \\ y''\Big|_{x = 1}=2 \\ 0=\frac d{dx}0=\frac d{dx}(2+2y'+xy''+6y(y')^2+3y^2y'')=3y''+xy'''+6(y')^3+18yy'y''+3y^2y''' \\ 0=3y''+6(y')^3+xy'''+y(18y'y''+3yy''') \\ 0=3y''\Big|_{x = 1}+6\left(y'\Big|_{x = 1}\right)^3+y'''\Big|_{x = 1}+\left(y\Big|_{x = 1}\right)\left(18y'\Big|_{x = 1}y''\Big|_{x = 1}+3y\Big|_{x = 1}y'''\Big|_{x = 1}\right) \\ 0=3\cdot2+6(-2)^3+y'''\Big|_{x = 1}=-42+y'''\Big|_{x = 1} \\ y'''\Big|_{x = 1}=42 $$