If $f'(a):\mathbb R^m\to \mathbb R^m$ is not an isomorphism, then Let $U\subset \mathbb R^m$ be an open set and $f:U\to \mathbb R^m$ a function of class $C^1$. Suppose there is  $a\in U$ such that $f'(a):\mathbb R^m\to \mathbb R^m$ is  not an isomorphism. then

$$\lim_{r\to 0}\frac{\operatorname{vol}f(B(a;r))}{\operatorname{vol}B(a;r)} = 0$$

I have previously shown that if $f'(a)$ for isomorphism, then $\lim_{r\to 0}\frac{\operatorname{vol}f(B(a;r))}{\operatorname{vol}B(a;r)} = |det f'(a)|$, the only thing I know, is that I can not use the change of variables theorem. 
Any suggestions on how to solve this problem?
 A: Your map $f'(a)$ is a linear map, represent it by a matrix $A$. There exist $U\in GL_m(\mathbb{C})$ such that $T=UA U^{-1}$ is upper triangular. Let $\lambda_1, \lambda_2, \dots,\lambda_m$ be the diagonal entries of $T$ (ie the eigenvalues of $A$).
$$ \varepsilon_0 := \min\{ \vert \lambda_i\vert : \lambda_i \neq 0 \}. $$ 
Then we have for $0 < \varepsilon < \varepsilon_0$, that the diagonal entries of 
$$ T + \varepsilon Id_m = U (A + \varepsilon Id_m )U^{-1} $$
are $\lambda_1 + \varepsilon, \dots , \lambda_m + \varepsilon$ (all of them are different from zero). Thus,
$$ det (A + \varepsilon Id_m) = \prod_{j=1}^m (\lambda_j + \varepsilon) \neq 0. $$
Set $g_\varepsilon(x)= \varepsilon x$, then we have
$$ (f+g_\varepsilon)'(a) = f'(a) + \varepsilon Id_m, $$
which is an isomorphism fo $0<\varepsilon < \varepsilon_0$. Hence, we have
$$ \lim_{r\rightarrow 0} \frac{vol \ f (B(a,r))}{vol \ B(a,r)} 
= \lim_{\varepsilon \rightarrow 0} \left( \lim_{r\rightarrow 0} \frac{vol \ f (B(a,r))}{vol \ B(a,r)} \pm\varepsilon  \right)
= \lim_{\varepsilon \rightarrow 0} \left( \lim_{r\rightarrow 0} \frac{vol \ f (B(a,r))}{vol \ B(a,r)} \pm \frac{vol \ \varepsilon Id_m (B(a,r))}{vol \ B(a,r)}  \right) 
$$
therefore, we have
$$ \lim_{r\rightarrow 0} \frac{vol \ f (B(a,r))}{vol \ B(a,r)} 
= \lim_{\varepsilon \rightarrow 0} \lim_{r\rightarrow 0} \frac{vol \ (f+ \varepsilon Id_m) (B(a,r))}{vol \ B(a,r)}
= \lim_{\varepsilon \rightarrow 0} \vert det(f'(a)+ \varepsilon Id_m) \vert
= 0. $$
Where we used your result and the fact that the determinant is a continuous function of the matrix.
