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Let's solve the given rational equation:

\[
\frac{3}{x} + \frac{x - 3}{x} = \frac{1}{x^2}.
\]

### Step 1: Simplify the left-hand side.
Both terms on the left have the same denominator \(x\), so we can combine them:

\[
\frac{3 + (x - 3)}{x} = \frac{1}{x^2}.
\]

Simplify the numerator:

\[
\frac{x}{x} = \frac{1}{x^2}.
\]

Since \(x \neq 0\) (to avoid division by zero), we can reduce \(\frac{x}{x}\) to 1:

\[
1 = \frac{1}{x^2}.
\]

### Step 2: Solve for \(x\).
Multiply both sides by \(x^2\) to eliminate the fraction:

\[
x^2 = 1.
\]

Now solve for \(x\):

\[
x = \pm 1.
\]

### Step 3: Verify the solutions.
- For \(x = 1\): The equation becomes \(\frac{3}{1} + \frac{1 - 3}{1} = \frac{1}{1^2}\), which simplifies to \(3 - 2 = 1\), which is true.
- For \(x = -1\): Substituting gives \(\frac{3}{-1} + \frac{-1 - 3}{-1} = \frac{1}{(-1)^2}\), which simplifies to \(-3 + 4 = 1\), which is also true.

Both solutions are valid.

### Final Answer:
\[
x = \pm 1.
\]

Thus, the correct answer is:

\[
\boxed{A} \quad x = \pm 1.
\]

---

Would you like more details on any step, or do you have further questions?  
Here are five related questions to explore:

1. How can we solve rational equations with more complicated denominators?
2. Why do we exclude \(x = 0\) from the domain in this problem?
3. What are some strategies to identify extraneous solutions in rational equations?
4. How can this technique be extended to systems of rational equations?
5. How do rational equations differ from polynomial equations?

**Tip:** Always check for restrictions on the variable to ensure no division by zero occurs.

Solve the rational equation for x: 3/x + (x - 3)/x = 1/x^2.

Learn how to solve the rational equation 3/x + (x - 3)/x = 1/x^2 by simplifying the expression and solving for x. This problem introduces rational equations and the quadratic formula, suitable for students in Grades 8-10.

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