Theorem
Let $f \colon U \to V$ be homeomorphism form open set U in normed space X, to open set V in normed set Y. Let $a\in U$, $b=f(a)$, $f'(x_0) \colon X \to Y$ exists and it is isomorphism. Then the derivative $(f^{-1})'(b)$ exists and $(f^{-1})'(b) = (f'(a))^{-1}.$
Proof in book
Let $g=f^{-1}$ It is \begin{equation*} f(a+x)-f(x)=f'(a)x+o(x) \end{equation*} We have to prove \begin{equation*} g(b+y)-f(b)-(f'(a))^{-1}y=o(y) \end{equation*} or \begin{equation*} \frac{g(y+b)-g(b)-(f'(a))^{-1}y}{||y||} \to 0 \ { when } \ y\to0. \end{equation*}
Set $x = g(y+b)-g(b)$ so
$x+a=g(y+b)$. We then have $f(x+a)=y+b$.
From this we obtain
\begin{equation} y=f(a+x)-f(a) \end{equation}
and so $g(y+b)-g(b)=(f'(a))^{-1}*o(x)$ so it is enough to show
$\frac{||x||}{||y||} $
is bounded but I don't know how.
I think you have mixed up the variables. And you are trying to divide in a normed space, which may not be possible. I assume you mean to divide by the $\textit{norm}$ of $y$.
I am picking up where you left off:
Note first that there is a $c>0$ such that $\frac{\|f'(x_0)(x-x_0)\|}{\|x-x_0\|}>c$ in a suitably small neighborhood of $x_0$ because $f'(x_0)$ is invertible.
Now, by the reverse triangle inequality,
$\underset{x\to x_0}\liminf\frac{\|f(x)-f(x_0)\|}{\|x-x_0\|}\ge\underset{x\to x_0}\liminf \left|\frac{\|f'(x_0)(x-x_0)\|}{\|x-x_0\|}-\frac{f(x)-f(x_0)-f'(x_0)(x-x_0)}{\|x-x_0\|}\right|=\underset{x\to x_0}\liminf \left|\frac{\|f'(x_0)(x-x_0)\|}{\|x-x_0\|}\right|\ge c\tag 1$
Now, use the continuity of $f,\ f^{-1}$ and $f'(x_0)^{-1}$ and apply the result in $(1)$:
$\underset{y\to y_0}\lim\frac{f^{-1}(y)-f^{-1}(y_0)-(f'(x_0))^{-1}(y-y_0)}{\|y-y_0\|}=\underset{x\to x_0}\lim\frac{x-x_0-(f'(x_0))^{-1}(f(x)-f(x_0))}{\|f(x)-f(x_0\|}=$
$\underset{x\to x_0}\lim\left(-f'(x_0)^{-1}\left(\frac{f(x)-f(x_0)-(f'(x_0)(x-x_0)}{\|x-x_0\|}\right)\frac{\|x-x_0\|}{\|f(x)-f(x_0)\|}\right)=0\tag 2$