Three. (25 points) As shown in Figure 4, given points $A(0,3), B(-2, -1), C(2,-1) \cdot P\left(t, t^{2}\right)$ is a moving point on the parabola $y=x^{2}$ located within $\triangle A B C$ (including the boundary), the line $B P$ intersects $A C$ at point $E$, and the line $C P$ intersects $A B$ at point $F$. Express $\frac{B F}{C E}$ as a function of the variable $t$.
Three, suppose the line $AB$ is the graph of the linear function $y=ax+b$. Substituting $A(0,3)$ and $B(-2,-1)$, we get
$$
3=b, -1=-2a+b \text{. }
$$

Solving these, we get $a=2, b=3$.
Therefore, the line $AB$ is the graph of the linear function $y=2x+3$. The x-coordinates of the intersection points of this line with the parabola $y=x^2$ satisfy the equation $x^2=2x+3$. Solving this, we get $x_1=-1, x_2=3$.

Similarly, the line $AC$ is the graph of the linear function $y=-2x+3$. The x-coordinates of the intersection points of this line with the parabola $y=x^2$ are $x_3=1, x_4=-3$.

Since $P(t, t^2)$ is a moving point on the parabola $y=x^2$ located inside $\triangle ABC$, we have $-1 \leqslant t \leqslant 1$.
Draw $MN \parallel BC$, then $BM=CN$. Thus,
$$
\begin{array}{c}
\frac{BF}{CE}=\frac{BF}{BM} \cdot \frac{CN}{CE}=\frac{BC}{BC-MP} \cdot \frac{BC-NP}{BC}=\frac{BC-NP}{BC-MP} \\
=\frac{BC-\left(\frac{1}{2}MN-t\right)}{BC-\left(\frac{1}{2}MN+t\right)}=\frac{2BC-MN+2t}{2BC-MN-2t}. \\
\text{ Also, } BC=4, \frac{MN}{BC}=\frac{3-t^2}{4}, MN=3-t^2, 2BC-MN \\
=8-\left(3-t^2\right)=t^2+5, \text{ therefore, } \\
\frac{BF}{CE}=\frac{t^2+2t+5}{t^2-2t+5}(-1 \leqslant t \leqslant 1) .
\end{array}
$$