1.6.11 Find all $f: \mathbf{R} \rightarrow \mathbf{R}$, such that for any $x, y \in \mathbf{R}$, we have
$$
f\left(x^{2}+y+f(y)\right)=2 y+f(x)^{2} .
$$
Parse: The necessary and sufficient condition for $f(a)=0$ is $a=0$.
From the condition, we know that there exists $a \in \mathbf{R}$ such that $f(a)=0$ (let $y=-\frac{f(x)^{2}}{2}$ in the equation). Taking $x=0, y=a$ in (1), we have $0=2 a+f(0)^{2}$. Also, taking $y=0$ in (1), we get $f^{2}(x)=f\left(x^{2}+f(0)\right)$. Replacing $x$ with $-x$ in this equation, we have $f^{2}(-x)=f\left(x^{2}+f(0)\right)$, i.e., $f^{2}(x)-f^{2}(-x)$, thus $f(-a)=0$. Taking $x=0, y=-a$ in (1), we have $0=-2 a+f(0)^{2}$. Combining $0=2 a+f(0)^{2}$, we get $2 f(0)^{2}=0, f(0)=0$, and hence $a=0$. Therefore, $f(a)=0 \leftrightarrow a=0$.
Next, prove that $f(x)$ is an odd function, and when $x>0$, $f(x)>0$.
In fact, taking $y=0$ in (1), we know $f\left(x^{2}\right)=f(x)^{2}$, so when $x>0$, $f(x)>0$. Now for $\alpha>0$, taking $(x, y)=\left(\sqrt{\alpha},-\frac{1}{2} f(\alpha)\right)$ in (1), we get
$$
f\left(\alpha-\frac{f(\alpha)}{2}+f\left(\frac{-f(\alpha)}{2}\right)\right)=-f(\alpha)+f(\sqrt{\alpha})^{2}=-f(\alpha)+f\left((\sqrt{\alpha})^{2}\right)=0 .
$$

From the previous proof, we know $\alpha=\frac{f(\alpha)}{2}-f\left(\frac{-f(\alpha)}{2}\right)$, thus,
$$
f(-\alpha)=f\left(-\frac{f(\alpha)}{2}+f\left(\frac{-f(\alpha)}{2}\right)\right) .
$$

In (1), taking $(x, y)=\left(0,-\frac{1}{2} f(\alpha)\right)$, we have
$$
f\left(-\frac{f(\alpha)}{2}+f\left(\frac{-f(\alpha)}{2}\right)\right)=-f(\alpha),
$$

Comparing (2) and (3), we know that $f(x)$ is an odd function.
Finally, prove that $f(x)=x$. From the previous proof, we only need to prove that for any $x>0$, $f(x)=x$. For any $\alpha>0$, taking $(x, y)=(\sqrt{\alpha},-\alpha)$ in (1), we have
$$
f(f(-\alpha))=-2 \alpha+f(\sqrt{\alpha})^{2}=-2 \alpha+f(\alpha),
$$

Thus, $2 \alpha=f(\alpha)-f(f(-\alpha))=f(\alpha)+f(f(\alpha))$. Therefore,
$$
f(2 \alpha)=f(f(\alpha)+f(f(\alpha)))=f\left(0^{2}+f(\alpha)+f(f(\alpha))\right)=2 f(\alpha)+f(0)^{2}=2 f(\alpha),
$$

Taking $(x, y)=(\sqrt{2 \alpha},-\alpha)$ in (1), we have
$$
f(2 \alpha-\alpha+f(-\alpha))=-2 \alpha+f(\sqrt{2 \alpha})^{2}=-2 \alpha \mid f(2 \alpha)=-2 \alpha+2 f(\alpha) .
$$

That is, $f(\alpha-f(\alpha))=-2(\alpha-f(\alpha))$. Let $\beta=\alpha-f(\alpha)$, then $f(\beta)=-2 \beta$. If $\beta>0$, then $-2 \beta<0$, which is a contradiction; if $\beta<0$, then $f(\beta)=-f(-\beta)<0$, which is also a contradiction. Therefore, $\beta=0$. That is, $f(\alpha)=\alpha$.

In summary, the only function that satisfies the conditions is $f(x)=x$ (it is obvious that $f(x)=x$ satisfies the conditions).