Step 1: Use the parametric rule.
With $x=x(t)$ and $y=y(t)$, $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$.
Step 2: Differentiate $x$.
From $x=\sinh^{-1}t+\log(t^2+1)$,
\[ \frac{dx}{dt}=\frac{1}{\sqrt{1+t^2}}+\frac{2t}{1+t^2}=\frac{\sqrt{1+t^2}+2t}{1+t^2}. \]
Step 3: Differentiate $y$.
From $y=\tan^{-1}t+\log|t|$,
\[ \frac{dy}{dt}=\frac{1}{1+t^2}+\frac{1}{t}=\frac{t^2+t+1}{t(1+t^2)}. \]
Step 4: Form the ratio.
\[ \frac{dy}{dx}=\frac{\dfrac{t^2+t+1}{t(1+t^2)}}{\dfrac{\sqrt{1+t^2}+2t}{1+t^2}}=\frac{t^2+t+1}{t(\sqrt{1+t^2}+2t)}. \]
Step 5: Expand the denominator.
\[ t(\sqrt{1+t^2}+2t)=t\sqrt{1+t^2}+2t^2. \]
Step 6: Tidy the surd.
Since $t\sqrt{1+t^2}=\sqrt{t^2(1+t^2)}=\sqrt{t^4+t^2}$,
\[ \frac{dy}{dx}=\frac{t^2+t+1}{2t^2+\sqrt{t^4+t^2}}. \]
\[ \boxed{\dfrac{t^2+t+1}{2t^2+\sqrt{t^4+t^2}}} \]