Answer
$y'''$ = $42$
Work Step by Step
if $x$ = $1$
then
$x^{2}+xy+y^{3}$ = $1$
$(1)^{2}+1(y)+y^{3}$ = $1$
$y^{3}+y$ = $0$
$y(y^{2}+1)$ = $0$
$y$ = $0$
Differentiating implicitly with respect to $x$
$2x+xy'+y+3y^{2}y'$ = $0$
Substituting $1$ for $x$ and $0$ for $y$
$2(1)+(1)y'+0+3(0)^{2}y'$ = $0$
$2+y'$ = $0$
$y'$ = $-2$
Differentiating
$2+xy''+y'+y'+3y^{2}y''+6yy'y'$ = $0$
$2+xy''+2y'+3y^{2}y''+6yy'y'$ = $0$
Substituting $1$ for $x$, $0$ for $y$ and $-2$ for $y'$
$2+(1)y''+2(-2)+3(0)^2y''+6(0)(-2)(-2)$ = $0$
$y''-2$ = $0$
$y''$ = $2$
Differentiating
$xy'''+y''+2y''+3y^{2}y'''+6yy'y''+6yy'y''+y'(6yy''+6y'y')$ = $0$
$xy'''+y''+2y''+3y^{2}y'''+6yy'y''+6yy'y''+6yy'y''+6y'y'y'$ = $0$
$xy'''+3y''+3y^{2}y'''+18yy'y''+6y'y'y'$ = $0$
Substituting $1$ for $x$, $0$ for $y$, $-2$ for $y'$ and $2$ for $y''$
$1y'''+3(2)+3(0)^2y'''+18(0)(-2)(2)+6(-2)(-2)(-2)$ = $0$
$y'''-42$ = $0$
$y'''$ = $42$