Answer
$\frac{x+y}{x^2+xy+y^2}$
Work Step by Step
Step 1. Factor the denominators: $x^3-y^3=(x-y)(x^2+xy+y^2)$ and $x^2-y^2=(x+y)(x-y)$
Step 2. Use the above results in the original expression, we have:
$\frac{\frac{1}{x^3-y^3}}{\frac{1}{x^2-y^2}}=\frac{\frac{1}{(x-y)(x^2+xy+y^2)}}{\frac{1}{(x+y)(x-y)}}=\frac{(x+y)(x-y)}{(x-y)(x^2+xy+y^2)}=\frac{x+y}{x^2+xy+y^2}$
