I've already asked before this question. Now I would like to prove the general case, i.e., I want to prove the equality when the $f'(a)$ is not an isomorphism. To sum up I want to prove this:
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. Show
$$\lim_{r\to 0}\frac{\operatorname{vol}f(B(a;r))}{\operatorname{vol}B(a;r)} = 0$$
My attempt 1
Using the inverse function theorem $f$ is not a local diffeomorphism since $f'(a)$ is not an isomorphism. (Is $f(B(a,r))$ a manifold?) if so, $f(B(a,r))$ has dimension strictly less than $m$, then it has measure zero and it must have empty interior. So we have $\operatorname{vol} f(B(a,r))=0$.
Am I on the right way?
Attempt 2
Another way is to follow the suggestion of John B. in the comments of his answer of my previous question:
"You can write $|\det f′(y)|\lt \epsilon$ and so you can estimate from above easily. From below since the volume is nonnegative we do have $\lim_{r\to 0}≥0$. The matter is instead that since $f′(a)$ is not an isomorphism one cannot use the change of variables formula... but the volume is still at most the integral on the right-hand side." John B.
In this case I couldn't use the fact $f'(a)\neq 0$ to prove the estimation he said in the first line.