Example 7 As shown in Figure 5, given points $A(0,3)$, $B(-2, -1)$, $C(2, -1)$, and $P\left(t, t^{2}\right)$ as 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 $t$.
Given that the line $AB$ is the graph of the linear function $y = ax + b$. Substituting $A(0,3)$ and $B(-2,-1)$, we solve to 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 $-1 \leq t \leq 1$.
Draw $MN \parallel BC$, then $BM = CN$. Thus,
$$
\begin{array}{l}
\frac{BF}{CE} = \frac{BF}{BM} \cdot \frac{CN}{CE} = \frac{BC}{BC - MP} \cdot \frac{BC - NP}{BC} \\
= \frac{BC - NP}{BC - MP} = \frac{BC - \left(\frac{1}{2} MN - t\right)}{BC - \left(\frac{1}{2} MN + t\right)}
\end{array}
$$
$$
= \frac{2BC - MN + 2t}{2BC - MN - 2t}.
$$

Since $BC = 4$ and $\frac{MN}{BC} = \frac{3 - t^2}{4}$, we have $MN = 3 - t^2$. Therefore, $2BC - MN = 8 - (3 - t^2) = t^2 + 5$. Hence, $\frac{BF}{CE} = \frac{t^2 + 2t + 5}{t^2 - 2t + 5} \ (-1 \leq t \leq 1)$.