If $\|\left(f'(x)\right)^{-1}\|\le 1 \Longrightarrow$ $f$ is an diffeomorphism

Clearly the operator norm $ \|f'(x)\| \geq 1 $, hence $ f'(x)$ is invertible for all $ x \in \mathbb{R}^n $, from existence theorem of ODEs there exists $ \lambda : [0,1] \rightarrow \mathbb{R}^n $ satisfying $$ \lambda'(t) = (f'(\lambda(t))^{-1}(b-f(a)),\ \ \ \lambda(0) = a $$ Hence we have the relation $$ \begin{align}f(\lambda(t)) = f(\lambda(0)) + \int^t_0 (f\circ\lambda)'(s)ds\\ = f(a) + \int^t_0 f'(\lambda(s))(\lambda'(s))ds\\ =f(a) + \int^t_0 (b-f(a))ds= (1-t)f(a) +tb \end{align}$$

I assume here what @smiley06 has proved in his answer. With it, I will prove that for every continuous curve $\beta:[0,1]\to\mathbb{R}^n$, there exist a unique $\lambda$ continuous such that $f(\lambda(t))=\beta(t)$. Indeed, let $\beta:[0,1]\to\mathbb{R}^n$ be a continuous curve. For each $t$ consider the segment of line $$\ell(t,r)=(1-r)\beta(0)+r\beta(t),\ r\in [0,1]$$

For each $t$, there exist (this is the proof of @smiley06) $\lambda (t,r)$ such that $f(\lambda(t,r))=l(t,r)$. Now define $\lambda (t)=\lambda (t,1)$ which implies that $$f(\lambda (t))=l(t,1)=\beta(t)$$

Assume now that $f(x)=f(y)$ and let $\alpha (t)=(1-t)x+ty$. Consider $F:[0,1]\times [0,1]\to\mathbb{R}^n$ defined by $$F(\eta,t)=(1-\eta)f(\alpha(t))+\eta f(y)$$

For each fixed $\eta$ let $G(\eta,t)$ be a curve (which exists as we have proved above) such that $f(G(\eta,t))=F(\eta,t)$ and $G(\eta,0)=x$. Note that $$f(G(\eta,1))=f(y),\ \forall\ \eta\in [0,1]\tag{1}$$

Because $f$ is a local diffeomorphim we conclude from $(1)$ that $G(\eta,1)=y$ for all $\eta\in [0,1]$ which implies that $y=x$.