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To solve parts (a) and (b) of this problem, we need to find the domain (or the set of valid inputs \(M_f\)) for the functions \( f(x) \).

### (a) \( f(x) = \frac{1}{x^2 + 2x - 3} \)

Here, the function involves a rational expression. For the function to be defined, the denominator cannot be zero. So, we need to find the values of \(x\) for which:

\[
x^2 + 2x - 3 = 0
\]

We solve this quadratic equation:

\[
x^2 + 2x - 3 = (x - 1)(x + 3) = 0
\]

Thus, \( x = 1 \) and \( x = -3 \). The function is undefined at these points.

**Domain for (a):**

\[
M_f = \mathbb{R} \setminus \{-3, 1\}
\]

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### (b) \( f(x) = \frac{1}{x - 3 + \sqrt{9 - x^2}} \)

Here, the function has two parts to consider for its domain:
1. The denominator \(x - 3 + \sqrt{9 - x^2} \neq 0\)
2. The square root \( \sqrt{9 - x^2} \) must have a non-negative argument, meaning \(9 - x^2 \geq 0\).

#### Condition 1: The denominator must not be zero.

\[
x - 3 + \sqrt{9 - x^2} \neq 0
\]

Rearranging gives:

\[
\sqrt{9 - x^2} \neq 3 - x
\]

We'll return to this after we analyze the second condition.

#### Condition 2: The square root must be defined.

\[
9 - x^2 \geq 0 \quad \Rightarrow \quad -3 \leq x \leq 3
\]

This gives a preliminary interval of possible values for \(x\), but we also need to ensure that the denominator isn't zero inside this interval.

Now, solving for when the denominator is zero:

\[
x - 3 + \sqrt{9 - x^2} = 0
\]

\[
\sqrt{9 - x^2} = 3 - x
\]

Squaring both sides:

\[
9 - x^2 = (3 - x)^2
\]

Expanding the right-hand side:

\[
9 - x^2 = 9 - 6x + x^2
\]

Simplifying:

\[
0 = -6x + 2x^2 \quad \Rightarrow \quad 2x^2 - 6x = 0
\]

Factoring:

\[
2x(x - 3) = 0
\]

So, \(x = 0\) or \(x = 3\). But, since the denominator is zero when \(x = 3\), we must exclude this value from the domain.

Thus, combining the conditions, the domain is:

**Domain for (b):**

\[
M_f = [-3, 3) \setminus \{3\}
\]

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Let me know if you have any questions or if you would like a detailed explanation for any part of this solution! 

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### Additional Questions:
1. What are the steps to solve quadratic equations in general?
2. How do you solve equations with square roots?
3. Can you explain how to factor quadratics like in part (a)?
4. What are common cases where a function becomes undefined?
5. How do you determine the domain of logarithmic and trigonometric functions?

### Tip:
Always check for restrictions in both the numerator and the denominator when working with rational functions to avoid undefined points.

This problem involves determining the domain of two functions: a rational function and a square root function. The solution covers key algebraic concepts such as solving quadratic equations and analyzing the domain of functions with square roots.

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