problem: Given the functions $f(x)=\dfrac{x+5}{3}$ and $g(x)=\dfrac{1}{f^{-1}(x)+1}$, find the value of $g(3)$.
solution: We begin by calculating the inverse function $f^{-1}(x)$. Substituting $ f^{-1}(x)$ into the function $f(x) = \frac{x + 5}{3}$, we get \[f(f^{-1}(x))=\dfrac{f^{-1}(x)+5}{3}.\]Since $f(f^{-1}(x)) = x$ for all $x$ in the domain of $f^{-1}$, we have \[x=\dfrac{f^{-1}(x)+5}{3}.\]Solving for $f^{-1}(x)$ gives $$f^{-1}(x)=3x-5.$$Therefore, we can rewrite $g(x)$ as $$g(x)=\dfrac{1}{3x-5+1}=\dfrac{1}{3x-4}.$$Then $$g(3)=\dfrac{1}{3 \cdot 3 - 4}=\boxed{\dfrac{1}{5}}.$$