$\textbf{My question:}$ Let $f:U\rightarrow R^n \in C^1$ with $U\subset R^n$ open. If $a$ is a singular point of $f$ such that $a\in B(a,\epsilon )$ and every point in $B(a,\epsilon )-\left\lbrace a \right\rbrace$ is not singular, then there exist a open set $a\in V\subset B(a,\epsilon )$ such that $f|_V$ is an homeomorphism in a open set $f(V)\subset R^n$.
$\textbf{My problems and attempts:}$ I'm not sure about if this result is true or false. This question arose for me while I was working in other problems. I thought to use the map $g:B(a,\epsilon )\rightarrow R^n$ defined by $g(x)=x+f(x)$ and try to prove that $a$ is not a singular value of $g$, then, for inverse function theorem, we would find an homeomorphism $g|_V $, where $V$ is the desired set. However, I do not get overcome this step and conclude the prove, even more, I am not either sure if $a$ is not a singular point of $g$...
Tangential remark, too long for a comment.
The point $a$ may be singular for $g$. Indeed, $Dg(a)=I+Df(a)$, where $I$ is the $n\times n$ identity matrix, so its eigenvalues are $$\{1, 1+\lambda_j\, :\ \lambda_j\text{ is a non-zero eigenvalue of }Df(a)\}. $$ Notice, indeed, that $Df(a)$ has $0$ as an eigenvalue; that's the definition of "singular point" for $f$. However, $Df(a)$ might have other eigenvalues $\lambda_j$ as well. One of these $\lambda_j$ may equal $-1$, in which case $Dg(a)$ is singular.
You can overcome this difficulty by letting $g(x)=Ax+f(x), $ where $A\ne -\lambda_j$ for all $j$. For this one, $a$ is not a singular point. However, the statement you want to prove is false, as the counterexample of doetoe shows; the function $x\in\mathbb R\mapsto x^2\in \mathbb R$.