Determine all functions \( f: \mathbf{R} \rightarrow \mathbf{R} \), such that for any \( x, y \in \mathbf{R} \), we have
\[
f(x) f(y f(x)-1)=x^{2} f(y)-f(x) .
\]
Solution: In (1), take $x=0$, we get
$$
f(0) f(y f(0)-1)=-f(0) \text {. }
$$

If $f(0) \neq 0$, then from the above equation, we get $f(x)=-1$, which does not satisfy (1).
Therefore, it must be that $f(0)=0$.
In (1), take $y=0$, we get
$$
f(x) f(-1)=-f(x) \text {, }
$$

i.e., $f(x)(f(-1)+1)=0$.
If $f(-1)+1 \neq 0$, then from the above equation, we get $f(x) \equiv 0$, which satisfies (1). Now consider the remaining case.
If $f(-1)+1=0$, then
$$
f(-1)=-1 \text {. }
$$

Suppose for some $a \in \mathbf{R}$, we have $f(a)=0$, then in (1), take $x=a, y=-1$, we get $a^{2} f(-1)=0$, so $a=0$, i.e., only when $a=0$, we have
$$
f(a)=0 .
$$

In (1), take $x=y=1$, we get
$$
f(1) f(f(1)-1)=0 \text {. }
$$

Therefore, $f(f(1)-1)=0$. Thus, $f(1)-1=0$,
i.e., $f(1)=1$.
In (1), take $x=1$, we get
$$
f(y-1)=f(y)-1 \text {. }
$$

From (1) and (5), we get
$$
f(x) f(y f(x))-f(x)=x^{2} f(y)-f(x),
$$

i.e., $f(x) f(y f(x))=x^{2} f(y)$.
In (6), take $x=-1$, we get
$$
-f(-y)=f(y) \text {. }
$$

In (6), take $y=1$, we get
$$
f(x) f(f(x))=x^{2} \text {. }
$$

Therefore, $f(y f(x)) \stackrel{(6)}{=} \frac{x^{2}}{f(x)} \cdot f(y)$
(8)
$$
=f(f(x)) f(y) \quad(\forall x \neq 0),
$$

i.e., $f(f(x) y)=f(f(x)) f(y) \quad(\forall x, y \in \mathbf{R})$.
In (6), let $x=y \neq 0$ (using (3)), we get
$$
f(x f(x))=x^{2} \quad(\forall x \neq 0) .
$$

Therefore, $f(x f(x))=x^{2} \quad(\forall x \in \mathbf{R})$.
From 『and $($, we get
the range of the function $f$ is $\mathbf{R}$ (i.e., $f(x)$ takes all values in $\mathbf{R}$).
From (11) and $\mathscr{Y}$, we get
$$
f(y z)=f(y) f(z) \quad(\forall y, z \in \mathbf{R}) .
$$

In (12), replace $z$ with $z+1$, we get
$$
f(y z+y)=f(y) f(z+1) \text {. }
$$

Then, from (5) and (12), we get
$$
\begin{array}{l}
f(y z+y)=f(y)(f(z)+1) \\
=f(y) f(z)+f(y)=f(y z)+f(y) .
\end{array}
$$

In the above equation, let $y \neq 0, z=\frac{x}{y}$, we get
$$
f(x+y)=f(x)+f(y) \quad(\forall y \neq 0) .
$$

Therefore, $f(x+y)=f(x)+f(y) \quad(\forall x, y \in \mathbf{R})$. (13)
In (10), replace $x$ with $x+1$, and use (13) and (8), we get
$$
\begin{array}{l}
f(x f(x)+f(x)+x+1)=x^{2}+2 x+1, \\
f(x f(x))+f(f(x))+f(x)+1=x^{2}+2 x+1, \\
x^{2}+\frac{x^{2}}{f(x)}+f(x)+1=x^{2}+2 x+1,
\end{array}
$$

i.e., $\frac{x^{2}}{f(x)}+f(x)=2 x \quad(x \neq 0)$.
Thus, $x^{2}+f^{2}(x)=2 x f(x) \quad(\forall x \in \mathbf{R})$,

i.e., $(x-f(x))^{2}=0$.
Therefore, $f(x)=x \quad(\forall x \in \mathbf{R})$.
In summary, the solutions to this problem are (the verification is obvious):
$$
f(x)=0 \quad(\forall x \in \mathbf{R}) \text { and } f(x)=x \quad(\forall x \in \mathbf{R}) .
$$