Let $A : \mathbb{R} \to M_n(\mathbb{R})$ be a smooth map. We have the following hypothesis
$$A(0) = 0, \qquad A'(0) = I_n, \qquad A''(0) = 0$$
and $A(t)$ is invertible for all $t \not= 0$. I would like to show that
$$A^{-1}(t) \sim \frac{1}{t} I_n \quad \mbox{as} \quad t\rightarrow 0$$
Here is what I began to do : using Taylor's expansion for $t\not= 0$, we have $$A(t) = t I_n + G(t)$$ where $\frac{G(t)}{t^2} \rightarrow 0$ as $t\rightarrow 0$. Therefore, for $t\not= 0$, $$I_n = t A^{-1}(t) + A^{-1}(t)G(t)$$ hence $$A^{-1}(t) = \frac{1}{t}I_n - A^{-1}(t)\bigg(\frac{1}{t}G(t)\bigg)$$ It remains to show that $$A^{-1}(t)\bigg(\frac{1}{t^2}G(t)\bigg) \rightarrow 0 \quad \mbox{as}~~t\rightarrow 0$$ What do you think about it ?
Let $A=[a_{i,j}]$. By Taylor, for every $i,j$,
$a_{i,j}(t)=a_{i,j}(0)+t{a_{i,j}}'(0)+t^2/2{a_{i,j}}"(0)+O(t^3)=t{a_{i,j}}'(0)+O(t^3)$
Thus $A(t)=tA'(0)+O(t^3)=t(I+O(t^2))$.
Finally, $A^{-1}(t)=1/t(I+O(t^2))^{-1}$; since when $||U||$ tends to $0$, $(I+U)^{-1}=I+O(||U||)$, we deduce that
$A^{-1}(t)=1/t(I+O(t^2))=1/tI+O(t)$.