jmc

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 $$f(f^{-1}(x))=\frac{f^{-1}(x)+5}{3}.$$Since $f(f^{-1}(x))=x$ for all $x$ in the domain of $f^{-1}$, we have $$x=\frac{f^{-1}(x)+5}{3}.$$Solving for $f^{-1}(x)$ gives $$f^{-1}(x)=3x-5.$$Therefore, we can rewrite $g(x)$ as $$g(x)=\frac{1}{3x-5+1}=\frac{1}{3x-4}.$$Then $$g(3)=\frac{1}{3\cdot 3-4}=\boxed{\frac{1}{5}}.$$