Six. (15 points) Given that $f(x)$ is a function defined on the set of real numbers $\mathbf{R}$, for any $x, y \in \mathbf{R}$, it always holds that
$$
f(x-f(y))=f(f(y))+x f(y)+f(x)-1 \text {. }
$$

Find $f(x)$, and calculate the value of $f(\sqrt{2014})$.
Six, let the range of the function $f(x)$ be $A \subseteq \mathbf{R}, f(0)=c$. Then $c \in A$.
For $x=y=0$, we get
$$
f(-c)=f(c)+c-1 \text {. }
$$

Therefore, $c \neq 0$.
Let $x=f(y)$, we get
$$
\begin{aligned}
f(0) & =f(x)+x^{2}+f(x)-1 \\
\Rightarrow f(x) & =\frac{c+1}{2}-\frac{x^{2}}{2} .
\end{aligned}
$$

Substituting equation (1) into the original equation, we get
Left side $=f(x-f(y))$
$$
\begin{array}{l}
=\frac{c+1}{2}-\frac{(x-f(y))^{2}}{2} \\
=\frac{c+1}{2}-\frac{x^{2}}{2}+x f(y)-\frac{f^{2}(y)}{2} \\
=f(x)+x f(y)-\frac{f^{2}(y)}{2},
\end{array}
$$

Right side $=\frac{c+1}{2}-\frac{f^{2}(y)}{2}+x f(y)+f(x)-1$.
Comparing both sides, we get
$$
\frac{c+1}{2}=1 \Rightarrow c=1 \text {. }
$$

Thus, for any $x \in \mathbf{R}$, we have
$$
f(x)=1-\frac{x^{2}}{2} \text {. }
$$

It is easy to verify that $f(x)=1-\frac{x^{2}}{2}$ satisfies the given conditions.
Therefore, the function we seek is $f(x)=1-\frac{x^{2}}{2}$.
When $x=\sqrt{2014}$,
$$
f(\sqrt{2014})=1-\frac{2014}{2}=-1006 .
$$