$$
\(\frac{1}{x}+\frac{1}{y}=3\)
$$


And we have two quantities to compare:
- Quantity A: $3 x y$
- Quantity B: $x+y$

Our goal is to determine the relationship between Quantity A and Quantity B. That is, is Quantity A greater than Quantity B, is Quantity B greater than Quantity A, are they equal, or is the relationship indeterminate?

Step 1: Combine the Given Equation
First, let's combine the left side of the given equation:

$$
\(\frac{1}{x}+\frac{1}{y}=\frac{y+x}{x y}=\frac{x+y}{x y}=3\)
$$


So, we have:

$$
\(\frac{x+y}{x y}=3\)
$$


Step 2: Express $x+y$ in Terms of $x y$
From the equation above, we can solve for $x+y$ :

$$
\(x+y=3 x y\)
$$


Now, let's look at our quantities:
- Quantity A: $3 x y$
- Quantity B: $x+y$

But from the equation we just derived, $x+y=3 x y$. So, substituting:


But from the equation we just derived, $x+y=3 x y$. So, substituting:
- Quantity B: $x+y=3 x y$

Now, comparing:
- Quantity A: $3 x y$
- Quantity B: $3 x y$

Step 3: Compare Quantity A and Quantity B
At this point, it seems that:
Quantity $\(\mathrm{A}=3 x y=\)$ Quantity B
So, Quantity A equals Quantity B.
But wait! We should consider if there are any restrictions or additional conditions that might affect this equality.

Step 4: Check for Restrictions or Additional Conditions
The original equation $\(\frac{1}{x}+\frac{1}{y}=3\)$ implies that neither $x$ nor $y$ can be zero (since division by zero is undefined). Other than that, $x$ and $y$ can be any real numbers (positive or negative) that satisfy the equation.

Given that $x+y=3 x y$, and we've established that $3 x y=x+y$, the equality holds based on the given equation.

However, let's test with actual numbers to ensure our conclusion is correct.

Step 5: Test with Numerical Values
Case 1: Let $x=1$

Solve for $y$ :

$$
\(\begin{gathered}
\frac{1}{1}+\frac{1}{y}=3 \\
1+\frac{1}{y}=3 \\
\frac{1}{y}=2 \\
y=\frac{1}{2}
\end{gathered}\)
$$


Now, calculate $3 x y$ and $x+y$ :
- $\(3 x y=3 \times 1 \times \frac{1}{2}=\frac{3}{2}\)$
- $\(x+y=1+\frac{1}{2}=\frac{3}{2}\)$

Here, Quantity A = Quantity B.
Case 2: Let $x=2$

Solve for $y$ :

$$
\(\begin{gathered}
\frac{1}{2}+\frac{1}{y}=3 \\
\frac{1}{y}=3-\frac{1}{2}=\frac{5}{2} \\
y=\frac{2}{5}
\end{gathered}\)
$$


Now, calculate $3 x y$ and $x+y$ :
- $\(3 x y=3 \times 2 \times \frac{2}{5}=\frac{12}{5}\)$
- $\(x+y=2+\frac{2}{5}=\frac{12}{5}\)$

Again, Quantity A = Quantity B.

Case 3: Let $x=-1$
Solve for $y$ :

$$
\(\begin{gathered}
\frac{1}{-1}+\frac{1}{y}=3 \\
-1+\frac{1}{y}=3 \\
\frac{1}{y}=4 \\
y=\frac{1}{4}
\end{gathered}\)
$$


Now, calculate $3 x y$ and $x+y$ :
- $\(3 x y=3 \times(-1) \times \frac{1}{4}=-\frac{3}{4}\)$
- $\(x+y=-1+\frac{1}{4}=-\frac{3}{4}\)$

Once more, Quantity A = Quantity B.
Step 6: General Proof
From the equation:

$$
\(\begin{gathered}
\frac{x+y}{x y}=3 \\
x+y=3 x y
\end{gathered}\)
$$


Thus:

$$
\(3 x y=x+y\)
$$


Therefore, Quantity A ( $3 x y$ ) is always equal to Quantity B ( $x+y$ ) under the given condition $\(\frac{1}{x}+$ $\frac{1}{y}=3\)$.

Step 7: Considering Edge Cases
Are there any values of $x$ and $y$ that satisfy the original equation where $\(x+y \neq 3 x y\)$ ?
From our algebraic manipulation, $\(x+y=3 x y\)$ is a direct consequence of $\(\frac{1}{x}+\frac{1}{y}=3\)$, so as long as the original equation holds, the equality between Quantity A and Quantity B must hold.

The only restrictions are $\(x \neq 0\)$ and $\(y \neq 0\)$, which are already implied by the original equation.
Step 8: Conclusion
After combining the given equation, testing with specific numerical values, and generalizing the relationship, we consistently find that:

Quantity A = Quantity B
Therefore, the correct relationship is that Quantity A is equal to Quantity B.
Final Answer
$C$ (Assuming the options are: A. Quantity A is greater, B. Quantity B is greater, C. The two quantities are equal, D. The relationship cannot be determined.)