A classical one. In the picture we have $f(A)+f(B)+f(C)=f(A)+f(C)+f(D)=0$ hence $f(B)=f(D)$. Indicating the length of the red segment by $k$ we find that $f(P)=f(Q)$ whenever $PQ=k$. Now choose a point $O$ as the origin of the plane. If $OQ\le 2k$ then we can choose a point $P$ such that $OP=PQ=k$, hence $f(O)=f(P)=f(Q)$. Hence $f$ is constant on $\overline{D}(O,2k)$. Now if $Q$ is any point, then we have a chain $P_1=O,\cdots, P_s=Q$ such that all distances $P_iP_{i+1}$ are at most $2k$. Hence $f(O)=f(Q)$ and $f$ is constant. Therefore $f\equiv 0$.

Attachments:

Claim 1 : If $A,B \in \mathbb{R}^2$ such that $\text{dist}(A,B) = \sqrt{3}$, then $f(A)=f(B)$. Proof : Construct distinct $C,D$ on opposite sides of $AB$, such that $CA=CB=DA=DB=1$, which clearly exist by triangle inequality. As $AB=\sqrt{3}$, we must have $CD=1$. Hence $f(A)+f(C)+f(D)=0$ and $f(B)+f(C)+f(D)=0$. Hence $f(A)=f(B)$ as claimed. $\square$ Claim 2 : If $A,B \in \mathbb{R}^2$ such that $\text{dist}(A,B) = 1$, then $f(A)=f(B)$. Proof : Construct $C$ such that $CA=CB=\sqrt{3}$, which clearly exists. By claim 1, $f(A)=f(C)=f(B)$. Hence $f(A)=f(B)$ as claimed. $\square$ To finish, for any $X \in \mathbb{R}^2$, consider $Y,Z$ such that $XY=YZ=ZX=1$, so that $f(X)+f(Y)+f(Z)=0$. By claim 2, $f(X)=f(Y)=f(Z)$. Hence $f(X)=0$. Hence, we are done!

Let $A,B,C\in \mathbb R^2$ such that they form an equilateral triangle with side length $1$. Reflect $A$ across $BC$ to get the rhombus $ABCA’$. Note that $f(A)+f(B)+f(C)=f(A’) +f(B)+f(C) =0$ so $f(A)=f(A’)$ and we get the following : For any two points $A,A’$ with $\operatorname{dist}(A,A’)=\sqrt{3}$ we have $f(A)=f(A’)$. Now take any point $A$, we know there exists a finite sequence of points $A=X_0,X_1,X_2,...,X_n=O$ starting from $A$ and ending at $O$ such that $\operatorname{dist} (X_i,X_{i+1})=\sqrt{3}$. So $f(A)=f(O)$ and from there we get $f(A)=0$. Remark To construct such a sequence we proceed as follows : Say $AO=r$ then if $r<\sqrt{3}$ draw the perpendicular bisector of $AO$ and then let $X_1$ be a point on it such that $AX_1=\sqrt{3}$ so the sequence $A,X_1,O$ works here. If $r>\sqrt{3}$ write $r=\lfloor r \rfloor+\{r\}$ and define $X_1,...,X_{\lfloor r \rfloor}$ such that $\operatorname{dist} (X_i,X_{i+1})=\sqrt{3}$ and $A,X_i,O$ are collinear. Now as $\{r\}<1<\sqrt{3}$ we can finish using the first case.

Suppose we have $2$ equilateral triangles with side length $1$, $\triangle ABC$ and $\triangle DBC$. Then by the given equality, we must have $f(A) = f(D)$. By this logic, any $2$ points that are a distance of $AD = \sqrt3$ from each other have the same value under $f$. Consider $2$ points less than $2\sqrt3$ distance from each other. We can construct an isosceles triangle with these $2$ points as the base and legs of length $\sqrt3$. These $2$ points must therefore have the same value under $f$. For $2$ points further than $2\sqrt3$ away, we can draw a chain of segments of length $\sqrt3$ from one point $A$ to the other point $B$ until one of the endpoints of the segments is less than $2\sqrt3$ distance away from $B$ and then do our isosceles triangle configuration. Thus, $f(X)$ is constant for any point $X$ so it always equals $0$.