$(x-2)^2 = 9$, so $x-2 = ±3$ ---> x = -1 and 5
$(y-3)^2 = 25$, so $y-3 = ±5$ ---> y = -2 and 8

You could calculate the values for every variation of $\frac{x}{y}$, which isn't a bad strategy since there's only 4:

x = -1, y = -2: $\frac{-1}{-2} = \frac{1}{2} = \frac{4}{8}$
x = -1, y = 8: $\frac{-1}{8}$
x = 5, y = -2: $\frac{5}{-2} = \frac{-20}{8}$
x = 5, y = 8: $\frac{5}{8}$

... but a way to save time is to consider what would result in the minimums and maximums. Since there is a negative value for both x and y, the minimum value of $\frac{x}{y}$ is the one that is MOST negative, i.e., using the value of x with the greatest absolute value to maximize the numerator. Vice versa for maximum value of x/y.

Either way, we find that the minimum value of x/y is $\frac{-20}{8}$ and the maximum value of x/y is $\frac{5}{8}$. The difference between them is 25/8.