[Example 1.6.3] Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that for all real numbers $x, y$ we have
$$
f(x)+f(y)=f(f(x) f(y)) .
$$
Since equation (*) holds for all real numbers $x, y$, we can explore by assigning some special values to $x, y$.
To find $f(x)$, we can look for a breakthrough using special values. For this purpose, we calculate $f(0)$.
Let $f(0)=a$.
Step 1: Let $x=y=0$, by $f(0)=a$, equation (*) becomes
$$
a+a=f(0)+f(0)=f(f(0) \cdot f(0))=f\left(a^{2}\right),
$$

Thus,
$$
f\left(a^{2}\right)=2 a \text {. }
$$

This leads us to substitute the special value $a^{2}$.
Step 2: Let $x=y=a^{2}$, by $f\left(a^{2}\right)=2 a$ we get
$$
4 a=f\left(a^{2}\right)+f\left(a^{2}\right)=f\left(f\left(a^{2}\right) f\left(a^{2}\right)\right)=f\left(4 a^{2}\right),
$$

Thus,
$$
f\left(4 a^{2}\right)=4 a .
$$

We then need to consider the special value $x=4 a^{2}$.
Step 3: Let $x=4 a^{2}, y=0$, equation (*) becomes
$$
f\left(4 a^{2}\right)+f(0)=f\left(f\left(4 a^{2}\right) \cdot f(0)\right)=f\left(4 a^{2}\right),
$$

Thus,
$$
f(0)=0 \text {. }
$$

Step 4: Let $x=x, y=0$, equation (*) becomes
$$
f(x)+f(0)=f(f(x) \cdot f(0))=f(0),
$$

Thus,
$$
f(x)=0 .
$$

Through the above four substitutions of special values, we have obtained the solution to this problem: $f(x)=0$.