Answer
$g(x) =\frac{x-1}{x+1}$
Work Step by Step
$f(x) = \frac{1}{x}$
$(\frac{f}{g}) (x) = \frac{x+1}{x^{2}-x}$
Let $y$ be the function $g(x)$
$\frac{\frac{1}{x}}{y} = \frac{x+1}{x^{2}-x}$
$\frac{1}{x} ({x^{2}-x}) = (x+1)y$
$\frac{\frac{1}{x} ({x^{2}-x})}{(x+1)} = y$
$\frac{\frac{1}{x} x({x-1})}{(x+1)} = y$
$\frac{x-1}{x+1} =y$
Therefore, $g(x) =\frac{x-1}{x+1}$
