Answer
$\frac{d^3y}{dx^3}=42$
Work Step by Step
$x^2+xy+y^3=1$
$2x+y+x\frac{dy}{dx}+3y^2\frac{dy}{dx}=0$
$\frac{dy}{dx}=\frac{-2x-y}{x+3y^2}$
$\frac{d^2y}{dx^2}=\frac{(x+3y^2)\frac{d}{dx}(-2x-y)+(2x+y)\frac{d}{dx}(x+3y^2)}{(x+3y^2)^2}$
$=\frac{(x+3y^2)(-2-\frac{dy}{dx})+(2x+y)(1+6y\frac{dy}{dx})}{(x+3y^2)^2}$
$=\frac{(x+3y^2)(-2+\frac{2x+y}{x+3y^2})+(2x+y)(1+6y\frac{-2x-y}{x+3y^2})}{(x+3y^2)^2}$
$=\frac{(x+3y^2)(\frac{-2x-6y^2+2x+y}{x+3y^2})+(2x+y)(\frac{x+3y^2-12yx-6y^2}{x+3y^2})}{(x+3y^2)^2}$
$=\frac{(x+3y^2)(-2x-6y^2+2x+y)+(2x+y)(x+3y^2-12yx-6y^2)}{(x+3y^2)}$
$=\frac{-2x^2-6xy^2+2x^2+xy-6xy^2-18y^4+6xy^2+3y^3+2x^2+6xy^2-24yx^2-12xy^2+xy+3y^3-12xy^2-6y^3}{(x+3y^2)}$
$=\frac{2x^2-24x^2y-24xy^2+2xy-18y^4}{(x+3y^2)}$
$\frac{d^3y}{dx^3}=\frac{\frac{d}{dx}(2x^2-24x^2y-24xy^2+2xy-18y^4)(x+3y^2)-(2x^2-24x^2y-24xy^2+2xy-18y^4)\frac{d}{dx}(x+3y^2)}{(x+3y^2)^2}$
$\frac{d^3y}{dx^3}=\frac{(4x-48xy-24x^2\frac{dy}{dx}-24y^2-48xy\frac{dy}{dx}+2xy-18y^4)(x+3y^2)-(2x^2-24x^2y-24xy^2+2xy-18y^4)(1+6y\frac{dy}{dx})}{(x+3y^2)^2}$
Substitude $x=1$, $y=0$, $\frac{dy}{dx}=-2$
$\frac{d^3y}{dx^3}=42$
