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: $$\frac{dy}{dt}=\frac{da(t)}{dt}x^2+\frac{db(t)}{dt}x+\frac{dc(t)}{dt}$$