Example $\mathbf{3}$ 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】We can set one of the variables to 0 to simplify the equation (since when a variable multiplies a function, setting the variable to 0 makes the whole expression 0, effectively ignoring the function).
Let \( x=0, f(0)=s \). Then
$$
f(s f(y)+y-1)=s+y-1 \text{. }
$$

To reduce the number of “\( f \)” in the original equation, wouldn't it be better to find where \( f=0 \)?
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
$$
\begin{array}{l}
f(0)=f(a)=0 \\
\Rightarrow s=0, a=s f(1-s)-s=0 .
\end{array}
$$

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