Find the acute angle y if \ 4y - \ 2y = 0.

We know that $\cos \theta = \sin (90^\circ - \theta)$.
So, we can rewrite $\cos 2y$ as $\sin (90^\circ - 2y)$.
Substituting this into the equation:

$$\sin 4y = \sin (90^\circ - 2y)$$

For acute angles, if $\sin A = \sin B$, then $A = B$.
Therefore,

$$4y = 90^\circ - 2y$$ $$4y + 2y = 90^\circ$$ $$6y = 90^\circ$$ $$y = \frac{90^\circ}{6}$$ $$y = 15^\circ$$

The acute angle $y$ is $\boxed{15^\circ}$.