The position of a particle moving in the $xy$-plane is given by the parametric equations $x(t)=\dfrac{6t}{t+1}$ and $y(t)=-\dfrac{8}{t^2+4}$. What is the slope of the line tangent to the path of the particle at the point where $t=2$ ?

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To find the slope of the line tangent, we need to find:

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$

Take the derivative of $x(t)$ and $y(t)$ using the quotient rule. We obtain:

$$\frac{dx}{dt}=x'(t)=\frac{6}{(t+1)^2}$$ $$\frac{dy}{dt}=y'(t)=\frac{16t}{(t^2+4)^2}$$

At $t=2$,

$$x'(2) = \frac{2}{3}$$ $$y'(2) = \frac{1}{2}$$ $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{2}{3}}{\frac{1}{2}} = \boxed{\frac{3}{4}}$$