Example 3 Let $f: \mathbf{R} \rightarrow \mathbf{R}$ satisfy
$$
\begin{array}{l}
f(x f(x)+f(x) f(y)+y-1) \\
=f(x f(x)+x y)+y-1 .
\end{array}
$$

Find the analytical expression of the function $f(x)$.
【Analysis】Let one of the variables be 0, which can simplify the equation (since when a variable multiplies a function, if the variable is 0, the entire expression becomes 0, effectively ignoring the function).
$$
\begin{array}{l}
\text { Let } x=0, f(0)=s. \text { Then } \\
f(s f(y)+y-1)=s+y-1 .
\end{array}
$$

In the original equation with more “$f$”, wouldn't it be better to find $f=0$?
Let $y=1-s$. Then
$$
f(s f(1-s)-s)=0,
$$

which means there exists $a$ such that $f(a)=0$.
Considering the many “$f$” in the original equation, let $x=a$. Then
$$
f(y-1)=f(a y)+y-1 \text {. }
$$

Let $y=1$. Then
$$
\begin{aligned}
f(0) & =f(a)=0 \\
\Rightarrow s & =0, a=s f(1-s)-s=0 .
\end{aligned}
$$

Substituting into equation (1) gives $f(y-1)=y-1$.
Therefore, $f(x)=x$.
“Substitution” is for “simplification”, to get $y=f(x)$, the assignment of $x$ should be reasonable. Multiplying a complex expression by 0 to make it “disappear” can greatly simplify the equation.