`f(x)={{:((x^(2)-9)/(x-3)",when "xne3),(k", when "x=3):}` at x = 3

Answer

To solve the problem, we need to find the value of \( k \) such that the function

\[ f(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & \text{when } x \neq 3 \\ k & \text{when } x = 3 \end{cases} \]

is continuous at \( x = 3 \).

### Step-by-step Solution:

1. **Understanding Continuity**:
A function is continuous at a point \( x = a \) if:
\[ \lim_{x \to a} f(x) = f(a) \]
In our case, we need to check if:
\[ \lim_{x \to 3} f(x) = f(3) = k \]

2. **Finding the Limit**:
We need to calculate \( \lim_{x \to 3} f(x) \) when \( x \neq 3 \):
\[ \lim_{x \to 3} \frac{x^2 - 9}{x - 3} \]
Notice that \( x^2 - 9 \) can be factored as:
\[ x^2 - 9 = (x - 3)(x + 3) \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} \]

3. **Canceling the Common Terms**:
Since \( x \neq 3 \), we can cancel \( (x - 3) \):
\[ \lim_{x \to 3} (x + 3) \]

4. **Evaluating the Limit**:
Now we can directly substitute \( x = 3 \):
\[ \lim_{x \to 3} (x + 3) = 3 + 3 = 6 \]

5. **Setting the Limit Equal to \( k \)**:
Since the function is continuous at \( x = 3 \), we set:
\[ k = \lim_{x \to 3} f(x) = 6 \]

### Conclusion:
Thus, the value of \( k \) is:
\[ \boxed{6} \]