Apply the three-part continuity test at x = -3

First check that $k(-3)$ is defined: from the piecewise definition, $k(-3) = -6$, so the first condition of continuity is satisfied. Next, compute the limit as $x \to -3$ using the formula for $x \neq -3$: factor $x^2 - 9 = (x-3)(x+3)$, so $\frac{x^2-9}{x+3} = x - 3$ for $x \neq -3$. Therefore $\lim_{x\to -3} k(x) = \lim_{x\to -3}(x-3) = -6$, so the second condition (limit exists) is also satisfied.

Now compare the limit to the function value: $\lim_{x\to -3} k(x) = -6$ and $k(-3) = -6$, so they are equal, meaning the third condition holds as well. Since all three conditions of the continuity definition are satisfied, $k(x)$ is continuous at $x = -3$, and there is no discontinuity to classify.