Example 1 Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y$, we have
$$
f(x) f(y)=f\left(x^{2}+y^{2}\right) .
$$
The constant functions $f(x)=1$ or $f(x)=0$ clearly satisfy the requirement. However, it is unknown whether there are other functions that meet the requirement.

Let $f$ satisfy the requirement. We hope to determine $f$ through (1) (this condition should be fully utilized). Let $x=y=0$, by (1) we get $f^{2}(0)=f(0)$, so $f(0)=0$ or 1.
If $f(0)=0$, then by (1)
$$
f\left(x^{2}\right)=f(x) f(0)=0,
$$

which means when $x \geqslant 0$, $f(x)=f\left((\sqrt{x})^{2}\right)=0$. Also, in (1) replace $y$ and $x$ with $-x$, we get
$$
f^{2}(-x)=f\left(2 x^{2}\right)=f\left((\sqrt{2} x)^{2}\right)=0,
$$

so $f(x)$ is the constant function 0.
If $f(0)=1$, then $f(x)=f(x) f(0)=f\left(x^{2}\right)$. We only need to consider the values of $f$ on the positive real numbers. At this time,
$$
f(x+y)=f\left((\sqrt{x})^{2}+(\sqrt{y})^{2}\right)=f(\sqrt{x}) f(\sqrt{y})=f(x) f(y) .
$$

In (2), let $y=x$, we get

Also,
$$
\begin{array}{c}
f(2 x)=f^{2}(x)=f\left(2 x^{2}\right), \\
f(2 x)=f\left((2 x)^{2}\right)=f\left(4 x^{2}\right),
\end{array}
$$

so $f\left(2 x^{2}\right)=f\left(4 x^{2}\right)$. Let $2 x^{2}=y$, then $f(y)=f(2 y)$ holds for all $y>0$. Therefore, $f^{2}(x)=f(2 x)=f(x), f(x)=0$ or 1.

If there exists some $y$ such that $f(y)=0$, then by (2), $f(x+y)=0$, i.e., for $x$ greater than $y$, $f(x)=0$. Since
$$
f\left(\frac{y}{2^{k}}\right)=f\left(\frac{y}{2^{k-1}}\right)=\cdots=f(y)=0,
$$

so for all $x>0, f(x)=0$.
Thus, the solutions to this problem are $f(x)=0$ or $f(x)=1$ or
$$
f(x)=\left\{\begin{array}{l}
0, \text { if } x \neq 0 ｝ \\
{1, \text { if } x=0 .}
\end{array}\right.
$$
(It is easy to verify that the last function also satisfies the condition.)