Let $f(x)=\frac{1}{x-3}$.  Find the largest $x$ that is not in the domain of $g(x)=f(f(x))$.
There are two ways for $x$ to not be in the domain of $g$: it can not be in the domain of $f$, or it can be in the domain of $f$ but not in the domain of $f\circ f$.  In the first case, the denominator of $f$ is zero, so
$$x-3=0\Rightarrow x=3.$$For the second case, we see that the denominator of $f(f(x))$ is $\frac{1}{x-3}-3$.  If this is zero, we have \[\frac{1}{x-3} = 3 \implies x-3 = \frac{1}{3} \implies x = 3+\frac13 = \frac{10}3.\]This is greater than $3$, so the largest $x$ not in the domain of $g$ is $\tfrac{10}{3}$.