Question 39095
Yes, do it the second way, where you had 
a^3 = xy and
a^4 = (x^2)(y^3)
Now solve the first for x and plug it in to the second
x = (a^3)/y and then
a^4 = [(a^6)/(y^2)](y^3) so that
y = a^(-2)
and since
a^3 = xy we have
x = a^5
Now from y = a^(-2) we have
a^2 = 1/y = y^(-1) and
x = a^5 = y^(-5/2)
and we're done.