$(I)$ Solution: From the given conditions, we have $\begin{cases} \frac{9}{4a^2} + \frac{6}{4b^2} = 1 \\ \frac{c}{a} = \frac{\sqrt{3}}{3} \\ a^2 = b^2 + c^2 \end{cases}$, solving these equations gives $a^2 = \frac{9}{2}, b^2 = 3$.

$\therefore$ The standard equation of the ellipse $C$ is $\frac{2x^2}{9} + \frac{y^2}{3} = 1$;

$(II)$ Proof: For $A(x_{1},y_{1})$, we have $2x_{1}^2 + 3y_{1}^2 = 9$, and $x_{1} \in \left[-\frac{3\sqrt{2}}{2}, \frac{3\sqrt{2}}{2}\right]$,

$|PA| = \sqrt{(x_{1}-1)^2 + y_{1}^2} = \sqrt{(x_{1}-1)^2 + 3 - \frac{2}{3}x_{1}^2} = \sqrt{\frac{1}{3}(x_{1}-3)^2 + 1}$,

For $B(x_{2},y_{2})$, similarly, we get $|PB| = \sqrt{\frac{1}{3}(x_{2}-3)^2 + 1}$, and $x_{2} \in \left[-\frac{3\sqrt{2}}{2}, \frac{3\sqrt{2}}{2}\right]$.

The function $y = \frac{1}{3}(x-3)^2 + 1$ is monotonic in the interval $\left[-\frac{3\sqrt{2}}{2}, \frac{3\sqrt{2}}{2}\right]$,

$\therefore$ we have $x_{1} = x_{2} \Leftrightarrow |PA| = |PB|$,

$\because x_{1} \neq x_{2}$, $\therefore |PA| \neq |PB|$,

$\therefore \triangle PAB$ cannot be an equilateral triangle.

Thus, the answers are:

$(I)$ The standard equation of the ellipse $C$ is $\boxed{\frac{2x^2}{9} + \frac{y^2}{3} = 1}$;

$(II)$ It is proven that $\triangle PAB$ cannot be an equilateral triangle, which is $\boxed{\text{proven}}$.