, 2017

We claim that $f(x, y)$ is a polynomial in $x$ and $y$ of degree at most $d$.
It is obvious that every such polynomial satisfies the desired condition. To prove the converse, let $f$ be a function satisfying the desired condition. Pick $(d+2)$ straight lines $\ell_1$, $\ell_2$, $\dots$, $\ell_{d+2}$ in $\mathbb{R}^2$ such that no two are parallel and no three are concurrent. Let the equation of $\ell_i$ be $h_i(x, y) = 0$, where $h_i$ is a linear polynomial.
For $i < j$, let $(a_{ij}, b_{ij})$ be the intersection of $\ell_i$ and $\ell_j$, and consider the polynomial
$$\varphi(x, y) = \sum_{1 \le i < j \le d+2} f(a_{ij}, b_{ij}) \prod_{\substack{k=1 \\ k \ne i, j}}^{d+2} \frac{h_k(x, y)}{h_k(a_{ij}, b_{ij})}.$$
It is easy to see that $\varphi(a_{ij}, b_{ij}) = f(a_{ij}, b_{ij})$ for all $i < j$, and that $\deg \varphi \le d$. We shall show that $\varphi(a, b) = f(a, b)$ for all $a, b \in \mathbb{R}$.
We first make an observation: if $\ell$ is a line such that $\varphi(a, b) = f(a, b)$ for at least $(d+1)$ points $(a, b)$ on $\ell$, then $\varphi(a, b) = f(a, b)$ for all points $(a, b)$ on $\ell$. Indeed, pick constants $A$, $B$, $C$ and $D$ such that $t \mapsto (At + B, Ct + D)$ parametrizes the line. Then, note that $\varphi(At + B, Ct + D) - f(At + B, Ct + D)$ is a polynomial of degree at most $d$, and that it vanishes at least $(d+1)$ points, so $\varphi(a, b) = f(a, b)$ for all $(a, b)$ on $\ell$.
Now, for a fixed $\ell_i$, note that $f(a_{ij}, b_{ij}) = \varphi(a_{ij}, b_{ij})$ for every $j \ne i$, so $f(a, b) = \varphi(a, b)$ for all points $(a, b)$ on $\ell_i$ from the claim. If $(c, d)$ is a point not lying on any $\ell_i$, then we can construct a line $\ell$ which passes through $(c, d)$, does not pass through any $(a_{ij}, b_{ij})$, and is not parallel to any $\ell_i$. Now $f(a, b) = \varphi(a, b)$ with $(a, b) = \ell_i \cap \ell$ for various $i$, so $f(a, b) = \varphi(a, b)$ for all $(a, b)$ on $\ell$. In particular, $\varphi(c, d) = f(c, d)$. This completes the proof. $\square$