$解:∵x+\frac{1}{x}=3,∴(x+\frac{1}{x})^2=9,即x^2+2+\frac{1}{x^2}=9.∴x^2+\frac{1}{x^2}=7.$
$∵(x+\frac 1x)^3=27,∴x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=27.∴x^3+\frac{1}{x^3}=18.$
$∵x^4+\frac{1}{x^4}=(x^2+\frac{1}{x^2})^2-2=47,$
$∴原式=\frac{18+7}{47+3}=\frac{1}{2}.$