178 Find all functions \( f: \mathbf{R} \rightarrow \mathbf{R} \) such that for all \( x, y \in \mathbf{R} \),
\[
f\left((x+y)^{2}\right)=(x+y)(f(x)+f(y)).
\]
Solution: Taking $x=y=0$ in equation (1), we get
$$
f(0)=0 \text {. }
$$

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

From equations (1) and (3), we have
$$
(x+y)(f(x)+f(y))=(x+y) f(x+y) \text {. }
$$

Thus, $f(x+y)=f(x)+f(y)(\forall(x+y) \neq 0)$.
From equation (3), it is easy to get
$$
-x f(-x)=f\left((-x)^{2}\right)=f\left(x^{2}\right)=x f(x) \text {. }
$$

Hence, $f(-x)=-f(x)(\forall x \neq 0)$.
From equations (5) and (2), we get
$$
f(-x)=-f(x)(\forall x \in \mathbf{R}) \text {. }
$$

From equations (4), (6), and (2), we have
$$
f(x+y)=f(x)+f(y)(\forall x, y \in \mathbf{R}) \text {. }
$$

From equations (3) and (7), we get
$$
\begin{array}{l}
(x+1)(f(x)+f(1))=(x+1) f(x+1) \\
=f\left((x+1)^{2}\right)=f\left(x^{2}+2 x+1\right) \\
=f\left(x^{2}\right)+2 f(x)+f(1)=x f(x)+2 f(x)+f(1),
\end{array}
$$

which simplifies to
$$
\begin{array}{l}
x f(x)+x f(1)+f(x)+f(1) \\
=x f(x)+2 f(x)+f(1) .
\end{array}
$$

Thus, $f(x)=x f(1)(\forall x \in \mathbf{R}, f(1)$ can be any real number $)$.
Clearly, the above expression satisfies all the conditions in the problem. Therefore, the function we seek is $f(x)=k x$ ( $\forall x \in \mathbf{R}, k$ is any real number).
(Wu Weizhao, School of Mathematics and Information Science, Guangzhou University,
$510006)$