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