jmc

There are three denominators in the formula for $f(x)$: $$x,1+\frac{1}{x},1+\frac{1}{1+\frac{1}{x}}.$$ For $f(x)$ to be undefined, one of these denominators must be $0$. We go through them one by one.

The simplest denominator, $x$, is $0$ if $x=0$.

The second denominator, $1+\frac{1}{x}$, is $0$ if $x=-1$.

The third denominator, $1+\frac{1}{1+\frac{1}{x}}$, is $0$ if $$\frac{1}{1+\frac{1}{x}}=-1.$$ We can solve like this: $$\begin{array}{rl}-1& =1+\frac{1}{x}\\ -2& =\frac{1}{x}\\ x& =-\frac{1}{2}\end{array}$$

Thus, the sum of the three points not in the domain of $f(x)$ is $0+(-1)+\left(-\frac{1}{2}\right)=\boxed{-\frac{3}{2}}$.