Let
$$f(x) = \frac{1}{1+\frac{2}{1+\frac 3x}}.$$
There are three real numbers $x$ that are not in the domain of $f(x)$. What is the sum of those three numbers?
The fraction in $f(x)$ becomes undefined if and only if the denominator $1+\frac 3x$ becomes 0. We have $1+\frac 3x = 0$ if and only if $\frac 3x = -1$ if and only if $x = \boxed{-3}$. (If $1 + \frac 3x = 0$, then $1 = -\frac 3x$, then $x = -\frac 3 1 = -3$.)
To find the values of $x$ that are not in the domain of $f(x)$, we need to consider the denominator of the function.
Notice that the denominator involves several layers of fractions: $\frac{1+\frac 3x}{1+\frac 2{1+\frac 3x}}$. We need to ensure that these fractions are not equal to zero, because dividing by zero is undefined.
To find out when each layer of fraction is zero, we set the numerator of each fraction to zero and solve for $x$.
First, consider $\frac 3x$. The numerator $3$ is a constant, so it is never equal to zero. Therefore, $\frac 3x$ is defined for all $x$ except where $x = 0$.
Next, consider $1+\frac 3x$. The numerator remains to be $3$, which is never equal to zero. So, $1+\frac 3x$ is defined for all $x$.
Finally, consider $1+\frac 2{1+\frac 3x}$. For this fraction to be defined, the denominator $1+\frac 3x$ must be nonzero. Setting $1+\frac 3x = 0$, we solve for $x$:
\begin{align*}
1+\frac 3x &= 0 \\
\frac 3x &= -1 \\
x &= -3.
\end{align*}
Therefore, $1+\frac 2{1+\frac 3x}$ is defined for all $x \neq -3$.
Now that we have determined the values of $x$ that are not in the domain of $f(x)$, we can conclude that the three values are $0$, $-3$, and all real numbers that make $1+\frac 2{1+\frac 3x} = 0$. However, we have already found that $1+\frac 2{1+\frac 3x}$ is defined for all $x \neq -3$, so there are no real numbers apart from $-3$ that make the denominator zero.
Therefore, the sum of the three numbers not in the domain of $f(x)$ is $0 + (-3) = \boxed{-3}$.
To find the numbers that are not in the domain of $f(x)$, we need to identify the values of $x$ that will cause the denominator of $f(x)$ to be equal to zero.
Let's find the denominator of $f(x)$ by simplifying the given expression.
Starting with the innermost fraction, we have:
$$1+\frac{3}{x}$$
To simplify further, we need a common denominator. Multiplying the first term by $\frac{x}{x}$, we have:
$$\frac{x}{x}+\frac{3}{x}$$
Combining the fractions, we get:
$$\frac{x+3}{x}$$
Now, substituting this into the numerator of the outer fraction, we have:
$$1+\frac{2}{\frac{x+3}{x}}$$
Simplifying the expression inside this fraction, we get:
$$1+\frac{2x}{x+3}$$
Now, substituting this into the denominator of $f(x)$, we have:
$$1+\frac{2}{1+\frac{2x}{x+3}}$$
To simplify this further, we need a common denominator. Multiplying the first term by $\frac{x+3}{x+3}$, we have:
$$\frac{x+3}{x+3}+\frac{2}{1+\frac{2x}{x+3}}$$
Combining the fractions, we get:
$$\frac{x+3+2}{x+3}$$
Simplifying the numerator, we get:
$$\frac{x+5}{x+3}$$
This is the denominator of $f(x)$. The values of $x$ that make the denominator zero are the numbers which make $x+3=0$. Solving this equation, we find:
$$x=-3$$
So, the only real number that is not in the domain of $f(x)$ is $x=-3$.
Therefore, the sum of the three numbers that are not in the domain of $f(x)$ is $-3$.