Useful (and often tested) fraction property: $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$

So we can rewrite and simplify our target expression as follows: $\frac{x + y}{xy}=\frac{x}{xy}+\frac{y}{xy}=\frac{1}{y}+\frac{1}{x}$

From here we must recognize that we can minimize the value of $\frac{1}{y}+\frac{1}{x}$ by maximizing the values of $x$ and $y$

Since $x$ and $y$ are integers and $3 < x < y < 9$, the maximum values $x$ and $y$ are $x=7$ and $y=8$

This means the minimum value of $\frac{1}{y}+\frac{1}{x}=\frac{1}{8}+\frac{1}{7}=\frac{7}{56}+\frac{8}{56}=\frac{15}{56}$

Answer: $\frac{15}{56}$