Example : (1) If $f : \mathbb{R}^2\rightarrow \mathbb{R}^2,\ f(x)=C\cdot x$ is dilation, then it is bi-Lipschitz map.
(2) More generally, we consider an inversion $f:\mathbb{R}^2-\{o\}\rightarrow \mathbb{R}^2-\{o\},\ (x)=\frac{x}{|x|^2}$. By a direct computation, we have $$ 0<\| Df\|^2\leq 2\cdot J\ \ast $$ at any point where $\|A\|=\sqrt{\sum_{i,j} \ A_{ij}^2}$ and $J=|{\rm det}\ Df|$.
If $g: \mathbb{R}^2\times \{0\}\rightarrow \mathbb{S}^2,\ g(x,y) = \frac{1}{1+r^2} (x,y,\frac{-1+r^2}{2})$, i.e. stereographic projection, then $g$ satisfies $\ast$.
(3) If $f=id: (M^n,g) \rightarrow (M^n,kg)$ where $k>0$, then $J=k^n=n^{-n/2}\|Df\|^n$.
Definition : $f:X\rightarrow Y$ between $n$-dimensional manifolds is a quasi-regular map if $ 0<\| Df\|^n\leq C\cdot J$ for global constant $C$.
Question : In the definition, any smooth map $f: X\rightarrow Y$ with $J\neq 0 $ is a quasi-regular map ?
SHORT ANSWER. The answer is negative. The mapping $f\colon (0, \infty)^2\to (0, \infty)^2$ defined as $f(x, y)=(x, \sqrt y)$ is smooth and not quasi-regular.
LONG ANSWER (Prior to edit).
If I understand it correctly, the answer is negative. Let $f\colon \mathbb R^n\to \mathbb R^n$ be a smooth function, let $x\in \mathbb R^n$ be fixed and denote $$ A:=Df(x).$$ Now $A$ is an $n\times n$ matrix, and the question asks whether $$\tag{1}\|A\|^n\le C|\det A|$$ holds for an absolute constant $C>0$, where $$ \|A\|:=\max \|Ax\|_{\mathbb R^n} / \|x\|_{\mathbb R^n}.$$ But (1) cannot hold, for the matrix $$ A_\epsilon:=\begin{bmatrix} 1 & 0 \\ 0 & \epsilon\end{bmatrix}$$ is such that $\|A_\epsilon\|=1$ and $\det A_\epsilon=\epsilon$, and $\epsilon$ can be made as small as we wish.