This is what I got. If there's anything wrong, please point it out.

g'(x)=[f'(f(f(x)))] [f'(f(x))] [f'(x)]

because f'(a)=b f(a)=a

therefore,

g'(a)= [f'(f(f(a)))] [f'(f(a))] [f'(a)]

= [f'(f(a))] [f'(a)]

= f['(a) ]

= f['(a) ]

= (b) (b) (b)

= b^3= (b) (b) (b)

= b^3