Let f(x) = 1/(x - 1) for x ≠ 1. The composite g(x) = f(f(x)) is formed by putting f(x) into f again…

Step 1. The inner function. $f(x)$ is not defined at $x = 1$, so $g$ is not defined there.

Step 2. The outer function. The outer $f$ cannot take the input $1$. So we need $f(x) \ne 1$. Solving $\dfrac{1}{x - 1} = 1$ gives $x = 2$.

Step 3. Simplify. For $x \ne 1$ and $x \ne 2$, $g(x) = \dfrac{1}{\frac{1}{x - 1} - 1} = \dfrac{x - 1}{2 - x}$. This rational function is continuous everywhere except where the denominator is zero, which is $x = 2$, and the point $x = 1$ was already removed.