Lets say I know $y(x_1), y'(x_1), y''(x_1), y(x_2), y'(x_2), y''(x_2)$ values exactly
I know only 2 points. What is the best way to find $y(x_3)$?
I don't know how I would include the second derivative into the midpoint rule. I also don't want to only use $y(x_2)$
$$y(x_3)\approx y(x_2)+(x_3-x_2)\cdot y'(x_2)+\frac{(x_3-x_2)^2}{2}\cdot y''(x_2).$$
My understanding is that there are ways of employing the data at $x_1$ also in such a framework --- perhaps somebody else might enlighten both of us.
Particularly if the $y''(x_2)\neq y''(x_3)$ then perhaps you could fit a cubic to the data you have at $x_2$ and $x_3$. I think in theory you could fit a degree five as you have six equations --- although sometimes the higher degree the polynomial fit, the more wibbly it gets... at least you would have a unique solution unlike degrees three and four.
You could also build up a finite difference table using the derivatives you have.
I think I like the degree five polynomial fit to be honest.
Write $$f(x)=ax^5+bx^4+cx^3+dx^2+ex+f.$$
Find the first two derivatives in terms of the parameters $a...f$.
Now input $x_1$ and $x_2$ into the six functions to generate six equations in six unknowns. Solve the resulting linear system and you have
$$y(x_3)\approx f(x_3).$$