The triangle with side lengths $3, 5$, and $k$ has area $6$ for two distinct values of $k$: $x$ and $y$. Compute $|x^2 -y^2|$.
1. Let the side lengths of the triangle be \(a = 3\), \(b = 5\), and \(c = k\). The area of the triangle is given as 6.
2. Using the formula for the area of a triangle with two sides and the included angle, we have:
   \[
   \text{Area} = \frac{1}{2}ab \sin \theta
   \]
   Substituting the given values:
   \[
   6 = \frac{1}{2} \cdot 3 \cdot 5 \cdot \sin \theta \implies 6 = \frac{15}{2} \sin \theta \implies \sin \theta = \frac{4}{5}
   \]
3. Since \(\sin \theta = \frac{4}{5}\), \(\theta\) could be in the first or second quadrant. Therefore, \(\cos \theta\) could be \(\frac{3}{5}\) or \(-\frac{3}{5}\).
4. Using the Law of Cosines to find \(k\):
   \[
   k^2 = a^2 + b^2 - 2ab \cos \theta
   \]
   Substituting \(a = 3\), \(b = 5\), and \(\cos \theta\):
   \[
   k^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cos \theta
   \]
5. For \(\cos \theta = \frac{3}{5}\):
   \[
   k^2 = 9 + 25 - 2 \cdot 3 \cdot 5 \cdot \frac{3}{5} = 34 - 18 = 16 \implies k = 4 \text{ or } k = -4
   \]
6. For \(\cos \theta = -\frac{3}{5}\):
   \[
   k^2 = 9 + 25 - 2 \cdot 3 \cdot 5 \cdot \left(-\frac{3}{5}\right) = 34 + 18 = 52 \implies k = \sqrt{52} \text{ or } k = -\sqrt{52}
   \]
7. The two distinct values of \(k\) are \(4\) and \(\sqrt{52}\). We need to compute \(|x^2 - y^2|\):
   \[
   |x^2 - y^2| = |16 - 52| = | -36 | = 36
   \]

The final answer is \(\boxed{36}\)