Three, (20 points) 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 $f\left(x^{2}+2 x y+y^{2}\right)=(x+y)(f(x)+f(y))$ and $|f(x)-k x| \leqslant\left|x^{2}-x\right|$.
Three, 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), we get
$$
f(0)=0.
$$

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

From equation (4), it is easy to see that
$$
-x f(-x)=f\left((-x)^{2}\right)=f\left(x^{2}\right)=x f(x).
$$

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

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 have $\frac{f\left(x^{2}\right)}{x^{2}}=\frac{f(x)}{x}$, thus,
$$
\frac{f(x)}{x}=\frac{f\left(x^{\frac{1}{2}}\right)}{x^{\frac{1}{2}}}=\frac{f\left(x^{\frac{1}{4}}\right)}{x^{\frac{1}{4}}}=\cdots=\frac{f\left(x^{\frac{1}{2^{n}}}\right)}{x^{\frac{1}{2^{n}}}},
$$

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

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

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