The function given is:

$f(x) = \frac{1}{\sqrt{7x^2 + 6x - 1}}$

To find the domain of this function, we need to make sure of two things:

1. The denominator must be non-zero, since division by zero is undefined.
2. The expression inside the square root must be positive, as the square root of a negative number is not real in the context of real numbers.

So, we must solve the inequality:

$7x^2 + 6x - 1 > 0$

Let's solve this inequality to determine the domain of $f(x)$. I will proceed with the calculations.The solution to the inequality $7x^2 + 6x - 1 > 0$ is:

$x \in (-\infty, -1) \cup \left(\frac{1}{7}, \infty\right)$

Thus, the domain of the function $f(x)$ is:

$(-\infty, -1) \cup \left(\frac{1}{7}, \infty\right)$

This is the interval where the function is defined.

Would you like further details or explanations?

Here are 5 related questions to deepen understanding:

1. How do we solve quadratic inequalities in general?
2. What happens if $7x^2 + 6x - 1 = 0$ instead of $> 0$?
3. How would the domain change if the function had no square root?
4. What is the significance of the square root in the denominator for determining the domain?
5. Could we apply a similar process to a cubic function?

Tip: When solving inequalities involving a quadratic, always find the critical points by solving the equation, and check intervals to determine where the expression is positive or negative.