$\underline{\textbf{Solution 1}}$

It is easy to see that $\angle{DCE} + \angle{ACB} = 180^\circ$. Hence: $$S_{CDE} = \frac{1}{2}\cdot DC \cdot CE \cdot \sin{DCE} = \frac{1}{2}\cdot 3\cdot 5\cdot \sin{ACB}=\frac{1}{2}\cdot 3 \cdot 5 \cdot \frac{AB}{AC} = \frac{1}{2}\cdot 3 \cdot 5 \cdot \frac{4}{5} =6 $$

$\underline{\textbf{Solution 2}}$

Draw $EH \perp BC$ and let the foot be $H$. It is easy to see that $\triangle{ABC}\cong \triangle{CHE}$. Hence $CH=AB=4$. It follows that $$S_{CDE}=\frac{1}{2}\cdot CD \cdot CH = \frac{1}{2}\times 3 \times 4 = 12$$