In triangle $ABC,$ $D$ and $E$ are on side $\overline{BC}$ such that $BD = DE = EC.$  If $x = \angle BAD,$ $y = \angle DAE,$ and $z = \angle EAC,$ then find
\[\frac{\sin (x + y) \sin (y + z)}{\sin x \sin z}.\]
By the Law of Sines in triangle $ABE,$
\[\frac{BE}{\sin (x + y)} = \frac{AE}{\sin B} \quad \Rightarrow \quad \sin (x + y) = \frac{BE \sin B}{AE}.\]By the Law of Sines in triangle $ADC,$
\[\frac{CD}{\sin (y + z)} = \frac{AD}{\sin C} \quad \Rightarrow \quad \sin (y + z) = \frac{CD \sin C}{AD}.\][asy]
unitsize (2 cm);

pair A, B, C, D, E;

B = (0,0);
D = (1,0);
E = (2,0);
C = (3,0);
A = (2.5,1.5);

draw(A--B--C--cycle);
draw(A--D);
draw(A--E);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, S);
label("$x$", A + (-0.75,-0.6));
label("$y$", A + (-0.35,-0.6));
label("$z$", A + (0,-0.5));
[/asy]

By the Law of Sines in triangle $ABD,$
\[\frac{BD}{\sin x} = \frac{AD}{\sin B} \quad \Rightarrow \quad \sin x = \frac{BD \sin B}{AD}.\]By the Law of Sines in triangle $AEC,$
\[\frac{CE}{\sin z} = \frac{AE}{\sin C} \quad \Rightarrow \quad \sin z = \frac{CE \sin C}{AE}.\]Hence,
\begin{align*}
\frac{\sin (x + y) \sin (y + z)}{\sin x \sin z} &= \frac{\frac{BE \sin B}{AE} \cdot \frac{CD \sin C}{AD}}{\frac{BD \sin B}{AD} \cdot \frac{CE \sin C}{AE}} \\
&= \frac{BE \cdot CD}{BD \cdot CE} \\
&= \frac{2BD \cdot 2CE}{BD \cdot CE} = \boxed{4}.
\end{align*}