For i) use that the arctan is the inverse function of the tan, both are monotone and the derivative of arctan is $$\frac{1}{1+x^2}$$ so it is diffferentiable.

for ii) you won't find a linear but an affine linear function, here a hint.
Try to scale your intervall to the right diameter and say that $f(a)$ should be $-1$ and $f(b)=1$

iii) yes. There is a rule how to find the derivative of the composition of two functions.

Critique on your proof:

At first you didn't proof the bijection very well, as you just say that $f^{-1}$ is a smooth function, here someone could say that this is circular, as you don't know if $f^{-1}$ exists at all.

In general don't make life to hard, you only need $C^1$ not $C^\infty$, here it is the same work, so never mind.

For your function $g$, it is a monotone increasing function and $f(a)=-\frac{1}{2}$ and $f(b)=\frac{1}{2}$ so that is not a bijection.

## Best Answer

The matrix $M=Df_a$ is of size $m\times n$ and $N=Df_a^{-1}$ of size $n\times m$. If $f$ is a diffeomorphism between the two spaces then the rule for taking derivative of compositions implies that $M N= 1_m$ (identity in ${\Bbb R}^m$) and $NM=1_n$. The rank, however, can not be bigger than the minimum value of $n$ and $m$ so $n=m$.