Let $U \subset \mathbb{R}^n$ an open and bounded set, and let $f:U \longrightarrow \mathbb{R}$ a continuous function such that, $\frac{\partial f(x)}{\partial x_i}$ exists for all $x \in U$ and $i = 1,2,\dots, n$. Suppose that, $\forall a \in \partial U = \{a \in \mathbb{R}^n \mid \forall \varepsilon>0, B_\varepsilon(a)\cap U \not= \emptyset \text{ and } B_\varepsilon(a)\cap (\mathbb{R}^n \setminus U) \not= \emptyset\}$, we have $$\lim_{x \to a}f(x) = 0.$$ So, there is $c \in U$, such that $\frac{\partial f(c)}{\partial x_i} = 0$ for $i = 1,2, \dots, n$.
This is what I've done:
Suppose that for all $c \in U$, there is $i_c \in \{1,2, \dots , n\}$, such that $\frac{\partial f(c)}{\partial x_{i_c}} \not = 0$. Without losing generality, let's suppose that $$\frac{\partial f(c)}{\partial x_{i_c}} > 0$$ So, there is $\delta_c > 0$ such that if $|t| < \delta_c$ $$f(c + te_{i_c}) > f(c)$$ ie, $c$ is a local minimum in the direction of $x_{i_c}$. But, I couldn't get any contradiction rs. Can anybody help me with this?