Ytukyg

I have 3 data points: $(x_1,y_1), (x_2,y_2), (x_3,y_3)$

$x_1,x_2,x_3$ are fixed values.
$y_1,y_2,y_3$ are all separate functions of a third variable, $t$

So:
$y_1=f_1(x,t)$
$y_2=f_2(x,t)$
$y_3=f_3(x,t)$

At any given $t$, I can fit a curve through these three data points (spline, Bezier or whatever):
$y=f(x)$, so $y$ and $frac{dy}{dx}$ can be calculated for any $x$.

What is $frac{dy}{dt}$ at any value of $x$?

The simplest way to fit the data through 3 points is to draw a quadratic $$y=ax^2+bx+c$$
In this case, $a$, $b$, and $c$ are also depending on a parameter $t$:
$$y(x,t)=a(t)x^2+b(t)x+c$$
You are not given $a(t)$, $b(t)$, and $c(t)$ directly, so you need to calculate them. What you have is a system of equations:
$$begin{cases}ax_1^2+bx_1+c=y_1=f_1(t)\ax_2^2+bx_2+c=y_1=f_2(t)\ax_3^2+bx_3+c=y_3=f_3(t)end{cases}$$
You can solve this in terms of unknowns $a$, $b$, $c$ using determinants:
$$a(t)=frac{begin{vmatrix}
f_1(t) & x_1 & 1 \
f_2(t) & x_2 & 1 \
f_3(t) & x_3 & 1
end{vmatrix}}{begin{vmatrix}
x_1^2 & x_1 & 1 \
x_2^2 & x_2 & 1 \
x_3^2 & x_3 & 1
end{vmatrix}}=frac{f_1(t)begin{vmatrix}
x_2 & 1 \
x_3 & 1
end{vmatrix}-f_2(t)begin{vmatrix}
x_1 & 1 \
x_3 & 1
end{vmatrix}+f_3(t)begin{vmatrix}
x_1 & 1 \
x_32 & 1
end{vmatrix}}{begin{vmatrix}
x_1^2 & x_1 & 1 \
x_2^2 & x_2 & 1 \
x_3^2 & x_3 & 1
end{vmatrix}}$$
You get a similar expression for $b(t)$ and $c(t)$.
Taking derivative with respect to time should be easy now:
$$frac{dy}{dt}=frac{da(t)}{dt}x^2+frac{db(t)}{dt}x+frac{dc(t)}{dt}$$