Does there exist a function $ f(x,y)$ of two real variables that takes natural numbers as its values and for which $ f(x,y)\equal{}f(y,z)$ implies $ x\equal{}y\equal{}z?$ 

[i]A. Hajnal[/i]
1. **Define the set of natural numbers and the power set:**
   Let \(\mathbb{N} = \mathbb{Z}_{>0}\) be the set of natural numbers. The power set of \(\mathbb{N}\), denoted \(2^{\mathbb{N}}\), is the set of all subsets of \(\mathbb{N}\).

2. **Establish a bijection between \(\mathbb{R}\) and \(2^{\mathbb{N}}\):**
   It is well known that there exists a bijection between the set of real numbers \(\mathbb{R}\) and the power set \(2^{\mathbb{N}}\). For each real number \(x\), let \(S_x\) denote the subset of \(\mathbb{N}\) associated with \(x\) under this bijection.

3. **Define the set \(A_x\) for each real number \(x\):**
   For each real number \(x\), define the set \(A_x\) as follows:
   \[
   A_x = \{2s : s \in S_x\} \cup \{2s + 1 : s \notin S_x\}.
   \]
   Note that \(A_x \subseteq \mathbb{N}\).

4. **Prove that \(A_x \neq A_y\) for distinct \(x\) and \(y\):**
   Suppose \(x \neq y\). Then \(S_x \neq S_y\). If \(A_x \subseteq A_y\), then by examining the even and odd elements separately, we would have \(S_x \subseteq S_y\) and \(\mathbb{N} \setminus S_x \subseteq \mathbb{N} \setminus S_y\), which implies \(S_x = S_y\), a contradiction. Therefore, \(A_x \not\subseteq A_y\).

5. **Construct the function \(f(x, y)\):**
   Define the function \(f\) as follows:
   \[
   f(x, x) = 1,
   \]
   and for \(x \neq y\), let \(f(x, y)\) be some element of
   \[
   1 + A_x \cap (\mathbb{N} \setminus A_y).
   \]
   This set is nonempty because \(A_x \not\subseteq A_y\). Note that this step requires the axiom of choice.

6. **Verify the properties of the function \(f\):**
   Suppose \(f(x, y) = f(y, z) = k\) and \(x \neq y\) and \(y \neq z\). Then,
   \[
   k - 1 \in [A_x \cap (\mathbb{N} \setminus A_y)] \cap [A_y \cap (\mathbb{N} \setminus A_z)].
   \]
   This intersection is clearly empty, leading to a contradiction. Therefore, \(x = y\) or \(y = z\). Without loss of generality, assume \(x = y\). Then \(f(x, y) = 1\), and \(f(a, b) = 1\) if and only if \(a = b\). Thus, \(y = z\), and consequently, \(x = y = z\).

This completes the construction and proof that such a function \(f(x, y)\) exists.

The final answer is \( \boxed{ f(x, y) } \) as constructed above.