This is a really neat result, good job finding this conjecture!

To prove this, it will be useful to remove as much algebra as possible. Note that, if the result is true for some cubic, it's true for all scalar multiples of that cubic, since stretching in the y-axis doesn't change the relevant geometry here. Thus it's enough to show this for cubics of the form x³+ax²+bx+c. But the result also remains true if we shift the cubic along the x axis. Hence, by substituting y = x-a/3 if necessary, it's actually enough to prove the result for the cubic f(x)=x³+ax+b.

Let the roots be p,q,r and let z=(p+q)/2 be the midpoint of p and q. Consider the two tangent line T to the curve at (z,f(z)), and consider the line L connecting the two points (z,f(z)) and (r,0). We want to show T and L are the same line, and it's clearly enough to show they have the same gradient. In other words, we want to prove f'(z)=f(z)/(z-r).

From Vieta's formulas, we know 2z+r=p+q+r=0. This r=-2z. If we substitute this in above, and multiply through by z-r = 3z, and rearrange a little, we see it's enough to show 8z³+2az=b.

Substitute back z=½(p+q). Then the left hand side becomes (p³+ap)+(q³+aq)+3pq(p+q), which is equal to (-b)+(-b)-3pqr. From Vieta's formulas again, we know pqr=-b, so the whole thing evaluates to b and the proof is complete.

Hopefully that makes sense! I'm not aware of any proof that totally circumvents the algebra, but I tried to keep it to a minimum.