Question 1 Given that $f(x)$ is a function defined on $\mathbf{R}$, for any $x, y \in \mathbf{R}$, it always holds that
$$
f(x-f(y))=f(f(y))+x f(y)+f(x)-1 .
$$

Find $f(x)$, and calculate the value of $f(\sqrt{2014})$. ${ }^{[1]}$
$(2014$, Beijing High School Mathematics Competition (Grade 10))
For any $y \in \mathbf{R}$, substituting $(f(y), y)$ into equation (1) yields
$$
\begin{array}{l}
f(0)=f(f(y))+f^{2}(y)+f(f(y))-1 \\
\Rightarrow f(f(y))=\frac{f(0)+1}{2}-\frac{f^{2}(y)}{2} .
\end{array}
$$

Substituting $(f(x), f(y))$ into equation (1) yields
$$
\begin{array}{l}
f(f(x)-f(y)) \\
=f(f(y))+f(x) f(y)+f(f(x))-1 .
\end{array}
$$

Substituting equation (3) into the above equation yields
$$
\begin{array}{l}
f(f(x)-f(y)) \\
=\frac{f(0)+1}{2}-\frac{f^{2}(y)}{2}+f(x) f(y)+ \\
\frac{f(0)+1}{2}-\frac{f^{2}(x)}{2}-1 \\
\Rightarrow f(f(x)-f(y)) \\
\quad=f(0)-\frac{(f(x)-f(y))^{2}}{2} .
\end{array}
$$

For any real number $x$, substituting $\left(\frac{x+1-f(f(0))}{f(0)}, 0\right)$ into equation (1) $(f(0) \neq 0)$ yields
$$
\begin{array}{l}
f\left(\frac{x+1-f(f(0))}{f(0)}-f(0)\right) \\
=f(f(0))+\frac{x+1-f(f(0))}{f(0)} f(0)+ \\
\quad f\left(\frac{x+1-f(f(0))}{f(0)}\right)-1 .
\end{array}
$$

Simplifying, we get
$$
\begin{array}{l}
x=f\left(\frac{x+1-f(f(0))}{f(0)}-f(0)\right)- \\
f\left(\frac{x+1-f(f(0))}{f(0)}\right) .
\end{array}
$$

Combining with equation (4), we have $f(x)=f(0)-\frac{x^{2}}{2}$ for any real number $x$.
Comparing with equation (3), we get $f(0)=1$.
Thus, $f(x)=1-\frac{x^{2}}{2}$.