<s>[INST] 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''
 [/INST] 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.
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