Calculus of a Single Variable, Chapter 2, 2.4, Section 2.4, Problem 93

We will need to use chain rule
(f(g(x)))'=f'(g(x))cdot g'(x)
First we find the first derivative.
f'(x)=-sin(x^2)cdot 2x=-2x sin(x^2)
Now we calculate second derivative, but for that we will need to use the following formula
(f cdot g)'=f' cdot g+f cdot g'
f''(x)=-2sin(x^2)-2xcos(x^2)cdot 2x
f''(x)=-2sin(x^2)-4x^2cos(x^2)
Now we evaluate the second derivative at 0.
f''(0)=-2sin(0^2)-4cdot0^2cos(0^2)=0
In the image below blue is the function and red is its second derivative.