#### Answer

$\frac{d^2y}{dx}=\frac{-1}{2}$

#### Work Step by Step

$y^3+y=2cosx$
Differentiate the above equation with respect to x:
$(3y^2+1)\frac{dy}{dx}=-2\sin{x}$
on taking double differentiation
$(3y^2+1)\frac{d^2y}{dx}+(6y)\frac{dy}{dx}=-2\cos{x}$
at point$(0,1)$
$(3+1)\frac{dy}{dx}=-2\sin{0}=0$
$\frac{dy}{dx}=0$
and
$(3+1)\frac{d^2y}{dx}+(6)\times0=-2\cos{0}$
$\frac{d^2y}{dx}=\frac{-1}{2}$