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''
 A: It's easier, I think, to avoid division as much a possible, thus:
With
$x^2 + xy + y^3 = 1, \tag 1$
and $x = 1$,
$1 + y + y^3 = 1; \tag 2$
thus
$y(y^2 + 1) = y^3 + y = 0; \tag 3$
thus, if we assume $y \in \Bbb R$,
$y = 0; \tag 4$
that is, we have
$(x, y) = (1, 0). \tag 5$
We find, taking $d/dx$ of (1),
$2x + y + xy' + 3y^2y' = 0; \tag 6$
with $x = 1$ and $y = 0$,
$2 + y'(1) = 0, \tag 7$
so 
$y'(1) = -2; \tag 8$
differentiating (6) yields
$2 + y' + y' + xy'' + 6y(y')^2 + 3y^2y'' = 0, \tag 9$
or
$2 + 2y' + xy'' + 6y(y')^2 + 3y^2 y'' = 0; \tag{10}$
we insert the values of $x, y, y'$:
$2 + 2(-2) + y'' = 0, \tag{11}$
whence
$y''(1) = 2; \tag{12}$
we now differentiate (10) one more time and obtain
$2y'' + y'' + xy''' + 6(y')^3 + 12yy'y'' + 6yy'y'' + 3y^2y''' = 0, \tag{13}$
or
$3y'' + xy''' + 6(y')^3 + 18yy'y'' + 3y^2y''' = 0; \tag{14}$
inserting the values for $x$, $y$, $y'$ and $y''$ yields
$3(2) + y''' + 6(-2)^3 = 0, \tag{15}$
or
$y''' = 42. \tag{16}$
This could obviously go on forever but I'll stop here.
A: 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  ?
