Let us consider a twice continuous differentiable function $f \colon \mathbb{R}^n \to \mathbb{R}$ such that
\begin{equation} \forall x \in \mathbb{R}^n : \ H_{f} (x) = \textbf{0}_{n \times n}. \end{equation}
Does it necessarily mean that $f$ must be a linear multivariable function, i.e., is the above mentioned criterion a characterization for continuous linear multivariable functions?
You need a vector-valued mean value theorem. The following will work:
Mean Value Theorem. Let $O$ be a open set in a normed space $V$ and $W$ a normed space. Let $f:O \to W$ be differentiable on $O$ and let $[x,y]$ be a segment contained in $O.$ If the derived function $f':O \to \mathscr{L}(O; W)$ (here $\mathscr{L}(V; W)$ is the space of all continuous linear functions $V \to W$) is bounded, meaning that $\| f'(x) \| \leq \lambda$ for all $x \in O$ (and $\|L\| = \sup\{Lx \mid \|x\| \leq 1\}$ for $L \in \mathscr{L}(O;W)$), then for any segment $[x,y] \subset O,$ we have $\|f(x) - f(y)| \leq \lambda \|x - y\|.$
The proof can be found in Dieudonné's Foundation of Modern Analysis chapter 8.
To your exercise. In three steps.
Step 1. I am going to show that if $f$ has zero derivative on an open connected $O,$ then $f$ is constant. Fix a base point $x_0 \in O$ and let $C$ be the set of points $x \in O$ such that $f(x) = f(x_0).$ Since $f$ is continuous, $C$ is closed. But $C$ is also open for if $x \in C$ and $B$ is a ball centred at $x$ small enough contained within $O,$ then the mean value theorem (with $\lambda = 0$), shows $f(y) = f(x)$ for all $y \in B,$ and since $f(x) = f(x_0)$ since $x \in C,$ we see $B \subset C.$ Connectedness gives $C = O.$
Step 2. If $O$ is open and connected and $f$ has constant derivative, then $f$ is the restriction of an affine function. To see this, consider $g = f - L,$ where $L$ is the unique linear function such that $f'(x) = L$ for all $x \in O.$ Clearly, $g' = 0,$ so $g$ is contant, therefore $c = g = f - L,$ so $f = L + c$ is affine.
Step 3. (End of proof.) Apply the previous two steps to $f,$ so $Df$ has zero derivative meaning it is constant, but if $Df$ is constant, then $f$ is an affine function itself. QED