Find the number of values of $x$ for which the expression $\frac{x^2-9}{x^2 + 9} + \frac{1}{x}$ is undefined.
We can see this expression has x as denominator. Hence, x = $\boxed{0}$.
To find the values of $x$ for which the expression is undefined, we need to look for two possibilities:
1) The denominator of any fraction becomes zero.
2) The value of $x$ makes the fraction $\frac{1}{x}$ undefined.
Let's analyze these possibilities step by step:
1) To find the values of $x$ for which the denominator of the first fraction, $(x^2 + 9)$, becomes zero:
$x^2 + 9 = 0$
Subtracting 9 from both sides:
$x^2 = -9$
Taking the square root of both sides (note that $\sqrt{-9}$ is undefined in the real number system):
$\sqrt{x^2} = \pm\sqrt{-9}$
Simplifying:
$x = \pm\sqrt{-9}$
Since $\sqrt{-9}$ is not a real number, there are no values of $x$ for which the denominator of the first fraction becomes zero.
2) To find the values of $x$ for which the fraction $\frac{1}{x}$ is undefined:
$\frac{1}{x}$ is undefined when $x = 0$, because division by zero is undefined.
Therefore, the expression $\frac{x^2-9}{x^2 + 9} + \frac{1}{x}$ is undefined when $x = 0$.
In summary, the expression is undefined for exactly one value of $x$, which is $x = 0$.