I know that for a function $f$ all of its partial derivatives are $0$.
Thus, $\frac{\partial f_i}{\partial x_j} = 0$ for any $i = 1, \dots, m$ and any $j = 1, \dots, n$.
Is there any easy way to prove that $f$ is constant? The results seems obvious but I'm having a hard time expressing it in words explicitly why it's true.
The result is FALSE in general, even in one dimension. Consider $f:(0,1) \cup (1,2) \to \mathbb{R}$ by $$f(x) = \left\{\begin{array}{lr} 0 & x \in (0,1)\\1 & x \in (1,2)\end{array} \right. $$ We have that $f'(x) = 0$ for all $x \in (0,1) \cup (2,3)$, but $f$ is not constant.
We do, however, have the following theorem which states that $f$ must be constant on the connected components of its domain:
Theorem. Let $U \subseteq \mathbb{R}^N$ be an open set, and suppose that $f:U \to \mathbb{R}$. If for every $x \in U$ and all $i \in \{ 1, \ldots, N\}$ there is $\frac{\partial f}{\partial x_i}(x) = 0$, then $f$ is constant on each connected component of $U$.
There are many proofs, and I will outline one here:
Proof.
Step 1: Prove that for every $x \in U$ there is a neighborhood of $x$ on which $f$ is constant. This result follows immediately from the Mean Value Theorem and the fact that $U$ is open.
Step 2: Let $K$ be a connected component of $U$. We know that $K$ is open since the connected components of an open set in $\mathbb{R}^N$ are open. Choose $x_0 \in K$, and define $V := f^{-1}(f(x_0))\cap K$. By step 1 we have that $V$ is open. Since $f$ is constant in a neighborhood of every point, it follows that $f$ is continuous. Hence $W:=f^{-1}(\mathbb{R}\setminus\{f(x_0)\}) \cap K$ is open (since $K$ is open). We note that $V \cup W = K$, and $V \cap W = \emptyset$. Since $x_0 \in V$ it follows that $V \neq \emptyset$. Hence we must have that $W = \emptyset$, otherwise $(V,W)$ would be a separation of $K$ which would be a contradiction since $K$ is connected. It follows that $K=V$ so that $f(x)=f(x_0)$ for all $x \in K$. $\blacksquare$
We assume that the function $f$ is everywhere defined and that the partial derivatives are identically $0$. Fix $i$. Because the partial derivative of $f_i$ with respect to $x_j$ is identically equal to $0$, the function $f_i$ does not vary as $x_j$ varies. This is true for all the $x_j$, so the function value is independent of the values of all the $x_j$.
But we took it for granted that if the (partial) derivative with respect to $x_j$ is $0$, the function value is independent of the value of $x_j$. We need to show this. It is really a one-variable problem. We want to show that if $g'(x)$ is identically $0$, then $g(x)$ does not depend on $x$.
This fact follows from the (one variable) Mean Value Theorem. By the MVT, for any $a$ and $b$ with $a \ne b$, there is a $c$ between $a$ and $b$ such that $$g(b)=g(a)+(b-a)g'(c).$$ But $g'(c)=0$, and therefore $g(a)=g(b)$.
It's not hard to see that given $\textbf{p,q}\in\mathbb{R}^M$:
$\hspace{2cm}\textbf{f(p)}-\textbf{f(q)}=\textbf{f}\bigl(f_1(\textbf{p})-f_1(\textbf{q}), f_2(\textbf{p})-f_2(\textbf{q}), ... f_N(\textbf{p})-f_N(\textbf{q})\bigr)$
$\hspace{4.5cm} =\bigl(\int_{\gamma} \nabla f_1(\mathbf{r})\cdot d\mathbf{r}, \int_{\gamma} \nabla f_2(\mathbf{r})\cdot d\mathbf{r}, ... \int_{\gamma} \nabla f_N(\mathbf{r})\cdot d\mathbf{r}\bigr)$
$\hspace{4.5cm}=0$
Since according to the Gradient Theorem, given any curve $\gamma$ with end points $\textbf{p},\textbf{q} \in \mathbb{R}^N$, we have:
$\hspace{6cm} f_i\left(\mathbf{p}\right)-f_i\left(\mathbf{q}\right) = \int_{\gamma} \nabla f_i(\mathbf{r})\cdot d\mathbf{r} $
and $\nabla f_i = \frac{\partial f_i}{\partial x_1 }\mathbf{e}_1 + \cdots + \frac{\partial f_i}{\partial x_N }\mathbf{e}_N=0$ for all $i$, by assumption.
