Let f and g be differentiable functions for all x\in\mathbb{R}

Now if f(x)=(g\circ f')(x), can I conclude that because f is differentiable,
thus (g\circ f') is differenitable.
And further that if (g\circ f') is differentiable then:
(g\circ f')'=g(f'(x))\cdot f''(x) and thus f is twice differentiable?

If not, then how about if g is bijective and thus has an inverse g^{-1}, and the inverse is differentiable.
Then:
f(x)=(g\circ f')
(g^{-1}\circ f)(x)=g^{-1}\circ(g\circ f')(x)=f'(x)

And now g^{-1}\circ f is a combination of 2 differentiable functions, thus
(g^{-1}\circ f)'=\left(f'(x)\right)'=f''(x)
then f is twice differentiable?

This is not homework, I am just thinking.