If $x+y+z=6$ and $xyz=2$, then find $\cfrac{1}{xy} +\cfrac{1}{yz}+\cfrac{1}{zx}$ 
If $x+y+z=6$ and $xyz=2$, then find the value of $$\cfrac{1}{xy}
 +\cfrac{1}{yz}+\cfrac{1}{zx}$$

I've started by simply looking for a form which involves the given known quantities ,so:
$$\cfrac{1}{xy} +\cfrac{1}{yz} +\cfrac{1}{zx}=\cfrac{yz\cdot zx +xy \cdot zx +xy \cdot yz}{(xyz)^2}$$
Now this might look nice since I know the value of the denominator but if I continue to work on the numerator I get looped :
$$\cfrac{yz\cdot zx +xy \cdot zx +xy \cdot yz}{(xyz)^2}=\cfrac{4\left(\cfrac{1}{xy}+\cfrac{1}{zy}+\cfrac{1}{zy}\right)}{(xyz)^2}=\cfrac{4\left(\cfrac{(\cdots)}{(xyz)^2}\right)}{(xyz)^2}$$
How do I deal with such continuous fraction ?

$$\cfrac{1}{xy} +\cfrac{1}{yz} +\cfrac{1}{zx}=\cfrac{yz\cdot zx +xy \cdot zx +xy \cdot yz}{(xyz)^2}=\frac{x+y+z}{xyz}$$
