The vertices (0,±4) are on the y-axis, and the center of the hyperbola will be midpoint of the line segment joining the two vertices. Thus, the center is (0,0). The standard form of the equation of a hyperbola with center at the origin (0,0) and vertices (0,±b) is given by
y2/b2 - x2/a2 = 1
where a and b are positive constants. Since the vertices in your problem are (0,±4), we have b = 4. Also, for the equation in the form above, it can be shown that the asymptotes of the hyperbola are given by
y = ±(b/a)x
Since the asymptotes in your problem are the lines y = ±(2/3)x, we can write
b/a = 2/3.
Now substitute b = 4 and solve the above for a.
4/a = 2/3
2a = 12
a = 6
Therefore, the equation of the hyperbola is
y2/42 - x2/62 = 1, or
y2/16 - x2/36 = 1
Hope that helps! Let me know if you need any further explanation.