Implicit Differentiation of this function [If 
$x^2 + xy + y^3 = 1$,
 find the value of 
$y'''$
 at the point where 
x = 1.]1
Am i going the right direction, if so how will know what is the values of y' and y''

First put $x=1$ into $x^2+xy+y^3=1$ to find $y$
$$x=1 \to 1+1y+y^3=1 \\y^3+y=0 \\y(y^2+1)=0 \\y=0\\(1,0)$$ your work point's is $(1,0)$
now 
$$x^2+xy+y^2=1 \to \\2x+1y+xy'+3y^2y'=0 \text{ put (1,0) }\\
2(1)+1(0)+y'+3(0)y'=0 \to y'=-2 $$ can you go on  ?
