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?
We must have $1+\frac 2 {1+\frac 3x} \neq 0$. Since the denominator is positive, this is equivalent to $1-\left| \frac 2 {1+\frac 3x} \right| \neq 0$. We can rewrite this as $\left| \frac 2 {1+\frac 3x} \right| \neq 1$. There are two cases: either $\frac 2 {1+\frac 3x} > 0$, or $\frac 2 {1+\frac 3x} < 0$.
Case 1: If $\frac 2 {1+\frac 3x} > 0$, then $\frac 2 {1+\frac 3x} \neq 1$ implies $\frac 2 {1+\frac 3x} > 1$. Solving this gives $\frac 3x < 1 \Longrightarrow x > 3$.
Case 2: If $\frac 2 {1+\frac 3x} < 0$, then $\frac 2 {1+\frac 3x} \neq 1$ implies $\frac 2 {1+\frac 3x} < -1$. Solving this gives $\frac 3x < -2 \Longrightarrow x > -\tfrac32$.
So we've found two numbers outside the domain so far: $x > 3$ and $x > -\tfrac32$. It remains to find a third number that will be excluded from the domain when we "smooth out" the absolute value signs. We consider the two possibilities $\frac 2 {1+\frac 3x} > 1$ and $\frac 2 {1+\frac 3x} < -1$.
Case 1: If $\frac 2 {1+\frac 3x} > 1$, then $1+\frac 3x > 2$. Solving, we find $x < 3$.
Case 2: If $\frac 2 {1+\frac 3x} < -1$, then $1+\frac 3x < -2$. Solving, we find $x > -\frac 14$.
We have $x > 3$, $x > -\tfrac32$, $x < 3$, and $x > -\tfrac 14$. The intervals that work for Case 1 are $(-\infty,-\tfrac 32)$, $(3,\infty)$, and the intervals that work for Case 2 are $\left(-\tfrac 32,-\tfrac 14\right)$ and $(3,\infty)$.
The only possible overlap is at $x=3$, so our three values are $\boxed{-\frac 32,-\frac 14,3}$.
To find the values of $x$ that are not in the domain of $f(x)$, we need to consider the denominators inside each fraction within the expression for $f(x)$.
Let's break it down step-by-step:
Step 1: Start with the innermost fraction $\frac 3x$. The denominator is $x$, so $x$ cannot be equal to $0$.
Step 2: The next fraction is $\frac{2}{1+\frac 3x}$. The denominator is $1+\frac 3x$. Since $x$ cannot be $0$ (as found in Step 1), we cannot have $\frac 3x$ equal to $-1$. Therefore, we have $1+\frac 3x\neq -1$, which means $x\neq -\frac 13$.
Step 3: Finally, we have the fraction $\frac{1}{1+\frac{2}{1+\frac 3x}}$. The denominator is $1+\frac{2}{1+\frac 3x}$. Again, $x$ cannot be $0$ (found in Step 1) and $x$ cannot be $-\frac 13$ (found in Step 2). Therefore, we must have $1+\frac 3x\neq -1$, $1+\frac 3x\neq -\frac 12$, and $1+\frac 3x\neq 0$.
To find the values of $x$ that satisfy these conditions, we can solve each inequality:
For $1+\frac 3x\neq -1$, we have $\frac 3x\neq -2$. Rearranging, we get $x\neq -\frac 32$.
For $1+\frac 3x\neq -\frac 12$, we have $\frac 3x\neq -\frac 32$. Rearranging, we get $x\neq -2$.
For $1+\frac 3x\neq 0$, we have $\frac 3x\neq -1$. Rearranging, we get $x\neq -3$.
Therefore, the three values of $x$ that are not in the domain of $f(x)$ are $-\frac 32$, $-2$, and $-3$.
To find the sum of these three numbers, we add them up:
$$-\frac 32 + (-2) + (-3) = -\frac 72 - 3 = -\frac{17}{2}.$$
To find the values of $x$ that are not in the domain of $f(x)$, we need to determine when the expression inside $f(x)$ becomes undefined.
Let's start by analyzing the expression inside $f(x)$:
$$1+\frac{3}{x}.$$
This expression will be undefined whenever the denominator, $x$, equals zero since division by zero is undefined.
So, we set $x$ equal to zero:
$$x = 0.$$
Therefore, $x = 0$ is one value that is not in the domain of $f(x)$.
Now, let's take a closer look at the expression involving $x$ in the denominator:
$$1+\frac{2}{1+\frac{3}{x}}.$$
Here, we need to ensure that the expression $1+\frac{3}{x}$ does not equal zero since division by zero is undefined.
To find the values of $x$ for which $1+\frac{3}{x}$ equals zero, we set the numerator equal to zero:
$$1 + 3/x = 0.$$
Solving this equation, we subtract 1 from both sides:
$$3/x = -1.$$
Then, we multiply both sides of the equation by $x$:
$$3 = -x.$$
Finally, we multiply both sides of the equation by $-1$ to solve for $x$:
$$x = -3.$$
Therefore, $x = -3$ is another value that is not in the domain of $f(x)$.
In summary, the two values not in the domain of $f(x)$ are $x = 0$ and $x = -3$. To find the third value, we need to consider the original expression of $f(x)$.
Since $f(x)$ is a composition of functions, $\frac{2}{1+\frac{3}{x}}$ needs to be defined first.
The expression $\frac{3}{x}$ will be undefined whenever $x$ equals zero since division by zero is undefined.
Therefore, in order for $\frac{2}{1+\frac{3}{x}}$ to be defined, $x$ cannot be zero.
So, the third value that is not in the domain of $f(x)$ is $x = 0$.
To find the sum of the three numbers not in the domain of $f(x)$, we add them together:
$$0 + (-3) + 0 = -3.$$
Hence, the sum of the three real numbers not in the domain of $f(x)$ is $-3$.