Suppose I have a smooth map $F:(x_1, x_2, x_3) \rightarrow (f_1(\mathbf{x}), f_2(\mathbf{x}),f_3(\mathbf{x}))$ from $\mathbb{R}^3$ to $\mathbb{R}^3$ and around a point $P= (P_1, P_2, P_3)$ the inverse function theorem applies. Suppose I take a small open box $B_{\varepsilon} = (P_1 - \varepsilon, P_1 + \varepsilon) \times (P_2 - \varepsilon, P_2 + \varepsilon) \times(P_3 - \varepsilon, P_3 + \varepsilon)$ so that the theorem applies.
I was just confused about something. If I take $\varepsilon$ very very small, does it then imply that image of $B_{\varepsilon}$ under $F$ also "look like" a box as well? or is it possible that it looks very differently? I understand my question is quite vague but I would appreciate any comments. And I will try to be more specific if anyone could ask me a question regarding this. Thank you.
If $F\colon \Bbb R^3 \to \Bbb R^3$ is smooth and $p_0 \in \Bbb R^3$ is such that ${\sf D}F(p_0)$ is non-singular, the inverse function theorem gives us two neighbourhoods $U$ of $p_0$ and $V$ of $F(p_0)$ such that $F\big|_U:U \to V$ is a diffeomorphism. You may choose $U$ or $V$ to be a box, but not both in general.
Since $U$ is open, you can fit a box $B$ around $p_0$, contained in $U$. Then $F\big|_B: B \to F[B]$ is a diffeomorphism, but $F[B]$ need not be a box.
Similarly, you can fit a box $B'$ around $F(p_0)$, contained in $V$. Then $F\big|_{F^{-1}[B']}\colon F^{-1}[B']\to B'$ is also a diffeomorphism, but $F^{-1}[B']$ need not be a box.
Note: you can write "open ball" instead of "box" if you want, and this remains valid in $\Bbb R^n$ in general.