Three. (50 points) For the bivariate polynomial $f(x, y)$ in terms of $x$ and $y$, it satisfies
$$
\begin{array}{l}
f(x, 0)=1, \\
f(f(x, y), z)=f(z, x y)+z .
\end{array}
$$

Find the expression for $f(x, y)$.
$$
\begin{array}{l}
f(1, y)=f(f(x, 0), y)=f(y, 0)+y=1+y . \\
\text { Then, } f(x, 1)=f(1+(x-1), 1) \\
=f(f(1, x-1), 1) \text { (by (1) } \\
=f(1,1 \cdot(x-1))+1 \text { (by the given) } \\
=[1+(x-1)]+1 \text { (by (1) } \\
=x+1 .
\end{array}
$$

On one hand, by (2) we have
$$
f(f(x, y), 1)=f(x, y)+1 \text {. }
$$

On the other hand, by (1) and the given formula, we also have
$$
f(f(x, y), 1)=f(1, x y)+1=(1+x y)+1 \text {. }
$$

Thus, $f(x, y)+1=(1+x y)+1$, which means
$$
f(x, y)=x y+1 \text {. }
$$

Verification shows that $f(x, y)=x y+1$ is the desired function.