Let $f: U \subset \mathbb{R}^{n} \to \mathbb{R}^{n}$ be a continuous function, where $U$ is open. Suppose that $f$ is differentiable in $x_{0} \in U$ and $D_{x_{0}}f$ is an isomorphism. Show that there is a neighborhood $V_{x_{0}} \subset U$ of $x_{0}$ such that $f(x) \neq f(x_{0})$ for any $x \in V_{x_{0}}\setminus\{x_{0}\}$.
My attempt My idea is suppose that in each neighborhood $V_{x_{0}}$ of $x_{0}$ there is $x \in V_{x_{0}}\setminus\{x_{0}\}$ such that $f(x) = f(x_{0})$ for get a contradiction.
We can take a sequence $(x_{n})$ with $x_{n} \in B_{1/n}(x_{0})$ such that $x_{n} \to x_{0}$ and $f(x_{n}) = f(x_{0})$. Write $x_{n} = x_{0} + t_{n}h_{n}$ where $h_{n}$ is a unity vector. Note that $$\Vert D_{x_{0}}f \Vert = \left\Vert \frac{f(x_{0}+t_{n}h_{n})-f(x_{0})}{t_{n}} - D_{x_{0}}f\right\Vert.$$
But, given $\epsilon > 0$, $$\left\Vert \frac{f(x_{0}+t_{n}h_{n})-f(x_{0})}{t_{n}} - D_{x_{0}}f\right\Vert < \epsilon$$ whenever $|t_{n}| < \delta$. Since $x_{n} \to x_{0}$, we can take $\delta>0$ such that $|t_{n}| < \delta$ for all $n > n_{0}$ for some $n_{0} \in \mathbb{N}$. For this $\delta$, we get $$\Vert D_{x_{0}}f \Vert < \epsilon,$$ then $D_{x_{0}}f$ cannot be an isomorphism.
Is this correct? The idea seems good to me, but I'm not sure about the details of the proof. Also, if its correct, I couldnt see where total continuity was necessary.
The equation $$\| D_{x_0}f \| = \left\| \frac{ f(x_0 + t_n h_n) - f(x_0)}{t_n} - D_{x_0}f \right\| $$ does not make sense to me. The quotient $\frac{f(x_0 + t_n h_n) - f(x_0)}{t_n}$ is a vector in $\mathbb{R}^n$, but $D_{x_0}f$ is an $n \times n$ matrix, so it is not meaningful to take their difference. Moreover, if the LHS refers to a matrix norm, there is no hope in proving that $ \| D_{x_0}f \|$ vanishes. (A singular matrix may still have large norm.) Instead, if you want to prove that $D_{x_0}f$ is not an isomorphism under your assumption, you should show that there is some vector $h \in \mathbb{R}^n \backslash\{0 \}$ so that $(D_{x_0}f) h$ (i.e. the vector multiplied by the matrix $D_{x_0}f$) is zero.
I think you should replace $D_{x_0}f$ by $(D_{x_0}f) h_n$ in all of the centered equations. If you do so, your work shows that there is a sequence $\{ h_n \}$ of unit vectors so that $(D_{x_0}f) h_n \to 0$. It follows by compactness of the unit ball that some subsequence $\{ h_{n_k} \}$ converges to a vector $h$. I claim that $(D_{x_0}f) h = 0$. Do you see why?