Example 9 Given a real number $k$, determine all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any $x, y \in \mathbf{R}$, we have $f\left(x^{2}+\right.$ $\left.2 x y+y^{2}\right)=(x+y)[f(x)+f(y)]$ and $|f(x)-k x| \leqslant\left|x^{2}-x\right|$.
From the problem, for any $x, y \in \mathbf{R}$, we have
$$
f\left[(x+y)^{2}\right]=(x+y)[f(x)+f(y)],
$$

and
$$
|f(x)-k x| \leqslant|x(x-1)| .
$$

Taking $x=y=0$ in equation (1) gives
$$
f(0)=0 .
$$

Taking $y=0$ in equation (1) and using equation (3) gives
$$
f\left(x^{2}\right)=x f(x) .
$$

From equation (4), we easily get
$$
-x f(-x)=f\left[(-x)^{2}\right]=f\left(x^{2}\right)=x f(x) .
$$

Thus, $f(-x)=-f(x)(x \neq 0)$.
Combining this with equation (3) gives
$$
f(-x)=-f(x) .
$$

From equation (5), we know that $f(x)$ is an odd function, so we only need to consider the case $x>0$.
From equation (4), we get $\frac{f\left(x^{2}\right)}{x^{2}}=\frac{f(x)}{x}$, hence,
$$
\frac{f(x)}{x}=\frac{f\left(x^{\frac{1}{4}}\right)}{x^{\frac{1}{2}}}=\frac{f\left(x^{\frac{1}{4}}\right)}{x^{\frac{1}{4}}}=\cdots=\frac{f\left(x^{\frac{1}{14}}\right)}{x^{\frac{1}{14}}},
$$

where $n \in \mathbf{N}^{+}$.
From equation (2), we get $\left|\frac{f(x)}{x}-k\right| \leqslant|x-1|$.
Taking $x$ as $x^{\frac{1}{2^{4}}}$ in equation (7) and using equation (6) gives
$$
\left|\frac{f(x)}{x}-k\right| \leqslant\left|x^{\Downarrow}-1\right| .
$$

Fixing $x$ in equation (8) and letting $n \rightarrow+\infty$, we get $x^{\frac{1}{2^{x}}} \rightarrow 1$, hence
$$
\left|\frac{f(x)}{x}-k\right| \leqslant|1-1|=0 .
$$

Thus, $\frac{f(x)}{x}-k=0$. Therefore, $f(x)=k x(x>0)$.
From equations (5) and (3), we get $f(x)=k x$.
Clearly, the above equation satisfies all the conditions in the problem.
Therefore, the solution to the problem is $f(x)=k x(x \in \mathbf{R})$.