Answer
$\frac{x-1}{x+1}, x\ne0$
Work Step by Step
$(f/g)(x)=f(x)/g(x)=\frac{1/x}{g(x)}=\frac{x+1}{x^2-x}$, thus $g(x)=\frac{x^2-x}{x(x+1)}=\frac{x-1}{x+1}, x\ne0$
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