Solution:
$(1)$ From the given conditions, we have $\begin{cases}e= \dfrac{c}{a}= \dfrac{\sqrt{6}}{3} \\ a^2=b^2+c^2 \\ \dfrac{1}{a^2}+ \dfrac{1}{b^2}=1\end{cases}$, solving these equations yields $\begin{cases}a^2=4 \\ b^2=\dfrac{4}{3}\end{cases}$,
$\therefore$ The equation of ellipse $C$ is $\dfrac{x^2}{4}+ \dfrac{3y^2}{4}=1$;
$(2)(i)$ When the slope of the tangent line $l$ of circle $O$ exists, let the equation of line $l$ be $y=kx+m$,
then the distance $d$ from the center $O$ to line $l$ is $d= \dfrac{|m|}{\sqrt{1+k^2}}$,
$\therefore 1+k^2=m^2$.
Combining the equation of line $l$ and the equation of ellipse $C$ $\begin{cases}y=kx+m \\ \dfrac{x^2}{4}+ \dfrac{3y^2}{4}=1\end{cases}$, we get $(1+3k^2)x^2+6kmx+3m^2-4=0$.
Let line $l$ intersect ellipse $C$ at points $A(x_1,y_1)$, $B(x_2,y_2)$,
then $x_1+x_2=- \dfrac{6km}{1+3k^2}$, $x_1x_2= \dfrac{3m^2-4}{1+3k^2}$.
$\therefore \vec{OA}\cdot \vec{OB}=x_1x_2+y_1y_2=x_1x_2+(kx_1+m)(kx_2+m)$
$=(1+k^2)x_1x_2+km(x_1+x_2)+m^2$
$=(1+k^2)\cdot \dfrac{3m^2-4}{1+3k^2}+km(- \dfrac{6km}{1+3k^2})+m^2$
$= \dfrac{4m^2-4-4k^2}{1+3k^2}$
$=0$,
$(ii)$ When the slope of the tangent line $l$ of the circle does not exist, it is verified that $\vec{OA}\cdot \vec{OB}=0$.
Combining the above, we conclude that $\vec{OA}\cdot \vec{OB}$ is a constant value $\boxed{0}$.