485 Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any real numbers $x, y$ the following holds:
$$
f\left(x^{2}+f(y)\right)=x f(x)+y .
$$
It is known that,
$$
f\left(x^{2}+f(y)\right)=x f(x)+y=-x f(-x)+y \text {. }
$$

Then for any $x \neq 0$, we have
$$
f(x)=-f(-x) \text {. }
$$

Thus, for $y \neq z$,
$$
\begin{array}{l}
f\left(x^{2}+f(y)\right)=x f(x)+y \\
\neq x f(x)+z=f\left(x^{2}+f(z)\right) .
\end{array}
$$

Therefore, $f(y) \neq f(z)$, which means $f(x)$ is injective.
Substituting $x=0, y=0$ into equation (1) respectively, we get
$$
\begin{array}{l}
f(f(y))=y, \\
f\left(x^{2}+f(0)\right)=x f(x) .
\end{array}
$$

Substituting $x=0, y=0$ into equation (1) again, we get $f(f(0))=0$.
Substituting $x=f(0)$ into equation (1), we get
$$
f\left(f^{2}(0)+f(y)\right)=y \text {. }
$$

Since $f(x)$ is injective, we have $f^{2}(0)=0$.
It is known that $f(x)$ is an odd function.
Substituting $f(0)=0$, we simplify to
$$
f(f(y))=y, f\left(x^{2}\right)=x f(x) \text {. }
$$

Notice that,
$$
\begin{array}{l}
f\left(x^{2}\right)=x f(x)=f(x) f(x)=f\left(f^{2}(x)\right) \\
\Rightarrow f^{2}(x)=x^{2} \\
\Rightarrow f(x)=x \text { or }-x .
\end{array}
$$

Clearly, conclusion (2) satisfies the problem.
If there exist $a, b \neq 0$ such that
$$
f(a)=a, f(b)=-b,
$$

then from $f\left(a^{2}+f(b)\right)=a f(a)+b$, we get
$$
f\left(a^{2}-b\right)=a^{2}+b \text {. }
$$

But $a^{2}-b \neq \pm\left(a^{2}+b\right)$, which is a contradiction.
Therefore, $f(x)=x$ or $f(x)=-x$.