Theorem.  If the function $f$ is defined in a domain $D$ of $\mathbb{R}^{n}$ and all the partial derivatives of a $f$ vanish identically in $D$, i.e.

then the function has a constant value in the whole domain.

Proof.  For the sake of simpler notations, think that  $n=3$; thus we have

Make the antithesis that there are the points  $P_{0}=(x_{0},\,y_{0},\,z_{0})$  and  $P_{1}=(x_{1},\,y_{1},\,z_{1})$  of $D$ such that  $f(x_{0},\,y_{0},\,z_{0})\neq f(x_{1},\,y_{1},\,z_{1})$.  Since $D$ is connected, one can form the broken line $P_{0}Q_{1}Q_{2}\ldots Q_{k}P_{1}$ contained in $D$.  When one now goes along this broken line from $P_{0}$ to $P_{1}$, one mets the first corner where the value of $f$ does not equal $f(x_{0},\,y_{0},\,z_{0})$.  Thus $D$ contains a line segment, the end points of which give unequal values to $f$.  When necessary, we change the notations such that this line segment is $P_{0}P_{1}$.  Now, $f_{x}^{{\prime}},\,f_{y}^{{\prime}},\,f_{z}^{{\prime}}$ are continuous in $D$ because they vanish.  The mean-value theorem for several variables guarantees an interior point  $(a,\,b,\,c)$  of the segment such that

But by (1), the last sum must vanish.  This contradictory result shows that the antithesis is wrong, which settles the proof.