Some details get lost because of the fact that $\displaystyle\lim_{x\to0}\frac{f(x)}{x}=1$. I have adjusted this answer to accept $\displaystyle\lim_{x\to0}\frac{f(x)}{x}=a$ to expose those details.
If $\displaystyle\lim_{x\to0}\frac{f(x)}{x}=a$, then $f'(f^{-1}(0))=f'(0)=a$. Furthermore, if $f^{(k)}(0)=0$ for $1<k<n$, but $f^{(n)}(0)\not=0$, we can use L'Hopital twice to get
$$
\begin{align}
\lim_{x\to0}\frac{f(x)-ax}{x/a-f^{-1}(x)}
&=\lim_{x\to0}\frac{f'(x)-a}{1/a-1/f'(f^{-1}(x))}\\
&=af'(f^{-1}(0))\;\lim_{x\to0}\frac{f'(x)-a}{f'(f^{-1}(x))-a}\\
&=a^2\cdot\lim_{x\to0}\frac{f''(x)}{f''(f^{-1}(x))/f'(f^{-1}(x))}\\
&=a^2f'(f^{-1}(0))\;\lim_{x\to0}\frac{f''(x)}{f''(f^{-1}(x))}\\
&=a^3\cdot\lim_{x\to0}\frac{f''(x)}{f''(f^{-1}(x))}\tag{1}
\end{align}
$$
Note that if $f^{(k)}(0)=0$, L'Hopital yields
$$
\begin{align}
a^{k+1}\cdot\lim_{x\to0}\frac{f^{(k)}(x)}{f^{(k)}(f^{-1}(x))}
&=a^{k+1}\cdot\lim_{x\to0}\frac{f^{(k+1)}(x)}{f^{(k+1)}(f^{-1}(x))/f'(f^{-1}(x))}\\
&=a^{k+1}f'(f^{-1}(0))\;\lim_{x\to0}\frac{f^{(k+1)}(x)}{f^{(k+1)}(f^{-1}(x))}\\
&=a^{k+2}\cdot\lim_{x\to0}\frac{f^{(k+1)}(x)}{f^{(k+1)}(f^{-1}(x))}\tag{2}
\end{align}
$$
Using $(1)$ and repeating $(2)$, we get that
$$
\lim_{x\to0}\frac{f(x)-ax}{x/a-f^{-1}(x)}=a^{n+1}\frac{f^{(n)}(0)}{f^{(n)}(f^{-1}(0))}=a^{n+1}\tag{3}
$$
Suppose $g^{(k)}(0)=0$ for $0\le k<m$ and $g^{(m)}(0)\not=0$ and $\displaystyle\lim_{x\to0}\frac{h(x)}{x}=1$. Then
$$
\begin{align}
\lim_{x\to0}\frac{g^{(k)}(h(x))}{g^{(k)}(x)}
&=\lim_{x\to0}\frac{g^{(k+1)}(h(x))h'(x)}{g^{(k+1)}(x)}\\
&=h'(0)\lim_{x\to0}\frac{g^{(k+1)}(h(x))}{g^{(k+1)}(x)}\\
&=\lim_{x\to0}\frac{g^{(k+1)}(h(x))}{g^{(k+1)}(x)}\\
&=\frac{g^{(m)}(h(0))}{g^{(m)}(0)}\\
&=1\tag{4}
\end{align}
$$
To finish things off, let $f(x)=\sin(\tan(\arcsin(\arctan(x))))$, $h(x)=\tan(\sin(x))$, and $a=1$. Then,
$$
\begin{align}
&\lim_{x\to0}\frac{\sin(\tan(x))-\tan(\sin(x))}{\arcsin(\arctan(x))-\arctan(\arcsin(x))}\\
&=\lim_{x\to0}\frac{f(h(x))-ah(x)}{h^{-1}(x/a)-h^{-1}(f^{-1}(x))}\\
&=\lim_{x\to0}\frac{f(h(x))-ah(x)}{h(x)/a-f^{-1}(h(x))}\frac{h(x)/a-f^{-1}(h(x))}{x/a-f^{-1}(x)}\frac{x/a-f^{-1}(x)}{h^{-1}(x/a)-h^{-1}(f^{-1}(x))}\\
&=a^{n+1}\tag{5}
\end{align}
$$
because
$$
\lim_{x\to0}\frac{f(h(x))-ah(x)}{h(x)/a-f^{-1}(h(x))}=a^{n+1}
$$
by $(3)$, and
$$
\lim_{x\to0}\frac{h(x)/a-f^{-1}(h(x))}{x/a-f^{-1}(x)}=1
$$
by $(4)$ using $g(x)=x/a-f^{-1}(x)$, and
$$
\lim_{x\to0}\frac{x/a-f^{-1}(x)}{h^{-1}(x/a)-h^{-1}(f^{-1}(x))}=\frac{1}{1/h'(h^{-1}(0))}=1
$$
Summary: Letting $g(x)=ah(x)$, we get that if
$$
\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}\frac{g(x)}{x}=a
$$
and $f^{(k)}(0)=g^{(k)}(0)$ for $1<k<n$, but $f^{(n)}(0)\not=g^{(n)}(0)$, then
$$
\lim_{x\to0}\frac{f(x)-g(x)}{g^{-1}(x)-f^{-1}(x)}=a^{n+1}
$$
Simpler approach: Convinced that there must be a simpler approach, I have revisited this answer.
For some $n>1$, assume
$f,g\in C^n$
$f^{(k)}(0)=g^{(k)}(0)$ for $k< n$ and $f^{(n)}(0)\not=g^{(n)}(0)$
$f(0)=0$ and $f'(0)=a\not=0$
These assumptions imply that $f(x)=ax+O(x^2)$ and $g(x)=ax+O(x^2)$.
Furthermore, $f^{-1}(x)=x/a+O(x^2)$ and $g^{-1}(x)=x/a+O(x^2)$.
Assumption 2 implies that
$$
\lim_{x\to0}\frac{f(x)-g(x)}{x^n}=\frac{f^{(n)}(0)-g^{(n)}(0)}{n!}\not=0\tag{6}
$$
Substituting $x\mapsto f^{-1}(x)$ and using $\lim\limits_{x\to0}\frac{g(g^{-1}(x))-g(f^{-1}(x))}{g^{-1}(x)-f^{-1}(x)}=g'(0)=a$ yields
$$
\begin{align}
\lim_{x\to0}\frac{f(x)-g(x)}{x^n}
&=\lim_{x\to0}\frac{f(f^{-1}(x))-g(f^{-1}(x))}{f^{-1}(x)^n}\\
&=\lim_{x\to0}\frac{g(g^{-1}(x))-g(f^{-1}(x))}{f^{-1}(x)^n}\\
&=\lim_{x\to0}\frac{a(g^{-1}(x)-f^{-1}(x))}{(x/a)^n}\\
&=a^{n+1}\lim_{x\to0}\frac{g^{-1}(x)-f^{-1}(x)}{x^n}\tag{7}
\end{align}
$$
Dividing both sides of $(7)$ by $\lim\limits_{x\to0}\frac{g^{-1}(x)-f^{-1}(x)}{x^n}$ yields
$$
\lim_{x\to0}\frac{f(x)-g(x)}{g^{-1}(x)-f^{-1}(x)}=a^{n+1}\tag{8}
$$
