解:
∵$\frac {xy}{x+y}=-2,$$\frac {yz}{y+z}=\frac {4}{3},$$\frac {zx}{z+x}=-\frac {4}{3},$
∴$\frac {x+y}{xy}=-\frac {1}{2},$$\frac {y+z}{yz}=\frac {3}{4},$$\frac {z+x}{zx}=-\frac {3}{4},$
$ $即$\frac {1}{x}+\frac {1}{y}=-\frac {1}{2},$$\frac {1}{y}+\frac {1}{z}=\frac {3}{4},$$\frac {1}{z}+\frac {1}{x}=-\frac {3}{4},$
三式相加得:
$ \begin {aligned}2(\frac {1}{x}+\frac {1}{y}+\frac {1}{z})&=-\frac {1}{2}+\frac {3}{4}-\frac {3}{4}\\\frac {1}{x}+\frac {1}{y}+\frac {1}{z}&=-\frac {1}{4}\end {aligned}$
∵$\frac {xy+yz+zx}{xyz}=\frac {1}{x}+\frac {1}{y}+\frac {1}{z}=-\frac {1}{4},$
∴$\frac {xyz}{xy+yz+zx}=-4。$