Solution: Clearly, $f(x)=x$ satisfies the requirements. Below, we prove that this is the only solution.
First, in (i), let $x=y=0$, we get
$$f(0)=f^{2}(0),$$

thus $f(0)=0$ or $f(0)=1$.
If $f(0)=1$, then for any $y \in \mathbf{R}$,
$$f(y)=f(0) f(y)=f(0)=1.$$

But in this case, $f(x+\sqrt{2})=f(x)=f(\sqrt{2})=1$, which contradicts (ii). Therefore, $f(0)=0$.
Similarly, in (i), let $x=y=1$, we get
$$f(1)=f^{2}(1),$$

thus $f(1)=1$ or $f(1)=0$.
If $f(1)=0$, then for any $-y \in \mathbf{R}$,
$$f(y)=f(1) f(y)=0,$$

which contradicts that $f(x)$ is not identically zero. Therefore, $f(1)=1$.
Next, we "improve" (ii). For any $y \neq 0$,
$$\begin{aligned} & f(x+y)=f\left(\frac{y}{\sqrt{2}}\left(x \cdot \frac{\sqrt{2}}{y}+\sqrt{2}\right)\right) \\ = & f\left(\frac{y}{\sqrt{2}}\right) f\left(x \cdot \frac{\sqrt{2}}{y}+\sqrt{2}\right) \\ = & f\left(\frac{y}{\sqrt{2}}\right)\left(f\left(x \cdot \frac{\sqrt{2}}{y}\right)+f(\sqrt{2})\right) \\ = & f\left(\frac{y}{\sqrt{2}}\right) f\left(x \cdot \frac{\sqrt{2}}{y}\right)+f\left(\frac{y}{\sqrt{2}}\right) f(\sqrt{2}) \\ = & f(x)+f(y) . \end{aligned}$$

The above equation clearly holds for $y=0$. Therefore, we have
(iii) $f(x+y)=f(x)+f(y)$.
Thus, $f(x)+f(-x)=f(0)=0$. Hence, $f(x)$ is an odd function, and we only need to consider $x>0$.
By $f(1)=1$ and (iii), it is easy to see that for $n \in \mathbf{N}$,
$$\begin{array}{l} f(n)=f(n-1)+f(1)=f(n-2)+2 f(1) \\ =\cdots=n f(1)=n . \end{array}$$

And for $m, n \in \mathbf{N}$, we have
$$\begin{array}{l} m f\left(\frac{n}{m}\right)=\underset{m \uparrow}{f\left(\frac{n}{m}\right)+f\left(\frac{n}{m}\right)+\cdots+f\left(\frac{n}{m}\right)} \\ =f \underbrace{\left(\frac{n}{m}+\frac{n}{m}+\cdots+\frac{n}{m}\right)}_{m \uparrow}=f(n)=n \text {, } \\ \end{array}$$

i.e.,
$$f\left(\frac{n}{m}\right)=\frac{n}{m} .$$

Thus, for all rational numbers $x$,
$$f(x)=x .$$

We only need to prove that this equation also holds for irrational $x$.
Since $f\left(x^{2}\right)=f(x) f(x)=f^{2}(x) \geqslant 0$, when $x>0$, $f(x)$ is non-negative. When $y>0$,
$$f(x+y)=f(x)+f(y) \geqslant f(x),$$

i.e., $f(x)$ is increasing.
For any irrational number $c$, we can find rational numbers $r_{1}$, $r_{2}$, such that $r_{1}<c<r_{2}$. By monotonicity,
$$r_{1}=f\left(r_{1}\right) \leqslant f(c) \leqslant f\left(r_{2}\right)=r_{2} .$$

Since $r_{1}, r_{2}$ can be arbitrarily close to $c$, we have
$$f(c)=c .$$

Thus, $f(x)=x$ holds for all $x$.
Note: After obtaining (iii), based on $f(1)=1$, we can deduce that (3) holds for all rational numbers $x$. This method is called the Cauchy method. However, to prove that (3) holds for all real numbers, (iii) alone is not sufficient; we must rely on monotonicity or continuity, which (i) provides.