11. A. As shown in Figure 4, the parabola $C: y=a x^{2}+b x-3$, with vertex $E$, intersects the $x$-axis at points $A$ and $B$, and the $y$-axis at point $C$. It is given that $O B=O C=3 O A$. The line $y=-\frac{1}{3} x+1$ intersects the $y$-axis at point $D$. Find $\angle D B C-\angle C B E$.
Three, 11. A.
It is easy to know $D(0,1), C(0,-3)$.
Then $B(3,0), A(-1,0)$.
Obviously, the line $y=-\frac{1}{3} x+1$ passes through point $B$.
Substitute the coordinates of point $C(0,-3)$ into
$$
y=a(x+1)(x-3),
$$

we get $a=1$.
Therefore, the vertex of the parabola $y=x^{2}-2 x-3$ is $E(1,-4)$.
By the Pythagorean theorem, we have
$$
\begin{array}{l}
B C=3 \sqrt{2}, C E=\sqrt{2}, B E=2 \sqrt{5}. \\
\text { Since } B C^{2}+C E^{2}=B E^{2} \\
\Rightarrow \angle B C E=90^{\circ} \\
\Rightarrow \tan \angle C B E=\frac{C E}{C B}=\frac{1}{3}.
\end{array}
$$

Also, $\tan \angle D B O=\frac{O D}{O B}=\frac{1}{3}$, so
$$
\begin{array}{l}
\angle D B O=\angle C B E. \\
\text { Therefore, } \angle D B C-\angle C B E \\
=\angle D B C-\angle D B O \\
=\angle O B C=45^{\circ}.
\end{array}
$$