Step 1: Build the hyperbola.
Centre at the origin with foci $(\pm3,0)$ means the transverse axis is along $x$, so $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ with $c=3$.
Step 2: Find $a$ from the eccentricity.
$e=\dfrac{c}{a}=\dfrac32$, so $\dfrac32=\dfrac{3}{a}$, giving $a=2$ and $a^2=4$.
Step 3: Find $b^2$.
$c^2=a^2+b^2\Rightarrow 9=4+b^2\Rightarrow b^2=5$. So the hyperbola is $\dfrac{x^2}{4}-\dfrac{y^2}{5}=1$.
Step 4: Substitute the line.
From $2x-y-1=0$, $y=2x-1$. Putting it in: $\dfrac{x^2}{4}-\dfrac{(2x-1)^2}{5}=1$.
Step 5: Form the quadratic.
Multiply by $20$: $5x^2-4(2x-1)^2=20$, that is $5x^2-16x^2+16x-4=20$, giving $11x^2-16x+24=0$.
Step 6: Test the discriminant.
$D=(-16)^2-4(11)(24)=256-1056=-800<0$. Since $D<0$ there are no real intersection points, so the line does not meet the hyperbola.
\[ \boxed{\text{does not intersect the hyperbola}} \]