Notation: $(x,y,z) \equiv (x^{1},x^{2},x^{3})$.
Consider a cartesian coordinate system and the euclidian vector space $\mathbb{E}^{3} \equiv (\mathbb{R}^{3},\ \langle \cdot, \cdot \rangle)$. Consider,also,the following proposition:
Let be $\Omega \subset \mathbb{R}^{3}$ a convex set and $\phi(x,y,z)$ a scalar field. If $$\vec{\nabla} \phi(x,y,z) = \vec{0}_{\mathbb{R}^{3}} = (0,0,0)$$ then $\phi(x,y,z) = constant$ in $\Omega$
My guess:
The gradient vector is then: $$\vec{\nabla} \phi(x,y,z) = \sum_{i=1}^{3}\frac{\partial \phi}{\partial x^{i}}\vec{e}_{i}$$
Where $\vec{e}_{i}$ are the canonical basis of $\mathbb{E}^{3}$ in cartesian coordinates, i.e. $$\beta = [\vec{e}_{1},\vec{e}_{2},\vec{e}_{3}] \equiv [\vec{i},\vec{j},\vec{k}]$$
Then,
$$\vec{\nabla} \phi(x,y,z) = \frac{\partial \phi}{\partial x}\vec{i} + \frac{\partial \phi}{\partial y}\vec{j} \frac{\partial \phi}{\partial z}\vec{k}$$
But,
$$\vec{\nabla} \phi(x,y,z) = \frac{\partial \phi}{\partial x}\vec{i} + \frac{\partial \phi}{\partial y}\vec{j} \frac{\partial \phi}{\partial z}\vec{k} = (0,0,0) $$
Then,
$$ \left\{ \begin{array}{c} \frac{\partial \phi}{\partial x}=0 \\ \frac{\partial \phi}{\partial y}=0 \\ \frac{\partial \phi}{\partial z}=0 \end{array} \right. $$
$$\frac{\partial \phi}{\partial x}=0 \implies \phi = \int dx + g(y,z) = 1 + g(y,z) = k^{1}$$
$$\frac{\partial \phi}{\partial y}=0 \implies \phi = \int dy + h(x,z) = 1+ h(x,z) = k^{2}$$
$$\frac{\partial \phi}{\partial z}=0 \implies \phi = \int dz + j(x,y) = 1 +j(x,y) = k^{3}$$
Then, $\phi$ is a constant field in $\Omega$.
Is my "proof" right?
Let $f(x,y,z) = \phi(x,y,z).$
$\nabla f • \nabla f = f_x^2 + f_y^2 + f_z^2 = 0.$
$f_x = f_y = f_z = 0$
Thus exists a,b,c with
$f(x,y,z) = f(a,y,z) = f(a,b,z) = f(a,b,c).$
The 1st equation because f is constant on x.
Likewise for the other equations.