Solution:

Because $N L \parallel A C$ we have triangles $D X L$ and $D E C$ are similar. From angle chasing, we also have that triangle $D E C$ is similar to triangle $A B C$. We have $\angle X N A = 180^{\circ} - \angle X N B = 180^{\circ} - \angle L N B = 180 - C A B = \angle L M A$. In addition, we have
$$\frac{N X}{N A} = \frac{X D \cdot X E}{X L \cdot N A} = \frac{A B}{B C} \frac{X E}{L C} \frac{N M}{N A} = \frac{A B}{B C} \frac{E D}{D C} \frac{B C}{A B} = \frac{E D}{D C} = \frac{A B}{A C} = \frac{M L}{M A}.$$
These two statements mean that triangles $A N X$ and $A M L$ are similar, and $\angle X A B = \angle X A N = \angle L A M = \angle L A C$. Similarly, $\angle X A Y = \angle L A C$, making $A, X$, and $Y$ collinear, with $\angle Y A B = \angle X A B = \angle L A C$; i.e., line $A X Y$ is a symmedian of triangle $A B C$.

Then
$$\frac{Z B}{Z C} = \frac{A B}{A C} \frac{\sin \angle Z A B}{\sin \angle Z A C} = \frac{A B}{A C} \frac{\sin \angle L A C}{\sin \angle L A B},$$
by the ratio lemma. But using the ratio lemma,
$$1 = \frac{L B}{L C} = \frac{A B}{A C} \frac{\sin \angle L A B}{\sin \angle L A C},$$
so
$$\frac{\sin \angle L A C}{\sin \angle L A B} = \frac{A B}{A C},$$
so
$$\frac{Z B}{Z C} = \frac{A B^2}{A C^2}.$$