The way you're trying to break up your function isn't quite right. The step where $x$ is multiplied with $\sqrt{1-x^2}$ should be seen as a product, not a composition. So you could look at this as $f(x)=g(x)\cdot h(x)$, where $h(x)=b(a(x))$ with $a$ and $b$ are as you defined them and $g(x)=x$. You can apply the product rule first, and only apply the chain rule for the part where you differentiate $h(x)$:
f'(x)=g'(x)h(x)+g(x)h'(x)=h(x)+xb'(a(x))a'(x)=\sqrt{1-x^2}+x\frac{1}{2\sqrt{1-x^2}}(-2x), etc. (You had a sign error in the derivative of $a$.)
The other problem is the notation where you wrote
$f(b(a))= \frac{1}{2~\sqrt{1-x^{2}}}~\cdot~2x= x \sqrt{1-x^{2}}.$
This doesn't really make sense. $f(b(a))$ is something different entirely from what you want to consider. The middle term is an attempted derivative of $b(a(x))$. And the right-hand term is $f(x)$. Thus, none of these are equal, and "$=$" is best reserved for equality, both to avoid errors and to allow meaningful communication to others (such as to whomever grades your assignment if this is for a class).