Step 1: Form $x-3y$.
With $x=\dfrac{t^3}{t^2-1}$ and $y=\dfrac{t}{t^2-1}$, $x-3y=\dfrac{t^3-3t}{t^2-1}=\dfrac{t(t^2-3)}{t^2-1}$.
Step 2: Differentiate $x$ with respect to $t$.
By the quotient rule, $\dfrac{dx}{dt}=\dfrac{(t^2-1)(3t^2)-t^3(2t)}{(t^2-1)^2}=\dfrac{t^4-3t^2}{(t^2-1)^2}=\dfrac{t^2(t^2-3)}{(t^2-1)^2}$.
Step 3: Set up the integral in $t$.
$\displaystyle\int\dfrac{dx}{x-3y}=\int\dfrac{\frac{t^2(t^2-3)}{(t^2-1)^2}}{\frac{t(t^2-3)}{t^2-1}}\,dt$.
Step 4: Cancel common factors.
The $(t^2-3)$ cancels, one power of $t$ cancels, and one $(t^2-1)$ cancels, leaving $\displaystyle\int\dfrac{t}{t^2-1}\,dt$.
Step 5: Substitute $u=t^2-1$.
Then $du=2t\,dt$, so $t\,dt=\dfrac{du}{2}$ and the integral becomes $\dfrac12\displaystyle\int\dfrac{du}{u}=\dfrac12\log|u|$.
Step 6: Return to $t$.
Substituting back, the answer is $\dfrac12\log(t^2-1)+C$.
\[ \boxed{\dfrac{1}{2}\log(t^2-1)+C} \]