Example 11 Let the function $f: \mathbf{R} \rightarrow \mathbf{R}$, for all $x, y \in \mathbf{R}$, satisfy $f\left[x^{2}+f(y)\right]=y+[f(x)]^{2}$.
Find $f(x)$.
Let $x=0$, and denote $f^{2}(0)=t$. From (1), we have
$$
f[f(y)]=y+t \text {. }
$$

From (2), we have $f\left[f\left(x^{2}+f(f(y))\right)\right]=x^{2}+f(f(y))+t$,
Noting (1), we have $f\left[f\left(x^{2}+f(f(y))\right)\right]=f\left[f(y)+(f(x))^{2}\right]$
$$
=f\left[(f(x))^{2}+f(y)\right]=y+[f(f(x))]^{2} .
$$

From (2), (3), and (4), we get $x^{2}+y+2 t=2 y+(x+1)^{2}$,
which simplifies to $2 t=t^{2}+2 t x$.
Thus, for any $x \in \mathbf{R}$, we have $t=0$, hence $f(0)=0$.
Therefore, (2) becomes $f[f(y)]=y$.
When $x \geqslant 0$, from (1) and (5), we get
$$
f(x+y)=f[x+f(f(y))]=f(y)+f^{2}(\sqrt{x}) \geqslant f(y) \text {, }
$$

Thus, $f(x)$ is an increasing function on $\mathbf{R}$, i.e., when $x \geqslant y$, we have $f(x) \geqslant f(y)$, which implies $f(x)=x$.
Indeed, if there exists $z \in \mathbf{R}$ such that $f(z) \neq z$, if $z>f(z)$, then $f(z) \leqslant f(f(z))=z$, a contradiction. If $z<f(z)$, then $f(z) \geqslant f(f(z))=z$, which is also a contradiction.