Answer
$$\frac{{xg\left( x \right)f'\left( x \right) + f\left( x \right)g\left( x \right) - xf\left( x \right)g'\left( x \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\left[ {\frac{{xf\left( x \right)}}{{g\left( x \right)}}} \right] \cr
& {\text{Let }}f\left( x \right){\text{ and }}g\left( x \right){\text{ differentiable functions with }}g\left( x \right) \ne 0 \cr
& {\text{use the quotient rule}} \cr
& \frac{d}{{dx}}\left[ {\frac{{xf\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{g\left( x \right)\frac{d}{{dx}}\left[ {xf\left( x \right)} \right] - xf\left( x \right)\frac{d}{{dx}}\left[ {g\left( x \right)} \right]}}{{{{\left[ {g\left( x \right)} \right]}^2}}} \cr
& {\text{use the product rule for }}\frac{d}{{dx}}\left[ {xf\left( x \right)} \right] \cr
& \frac{d}{{dx}}\left[ {\frac{{xf\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{g\left( x \right)\left( {x\frac{d}{{dx}}\left[ {f\left( x \right)} \right] + f\left( x \right)\frac{d}{{dx}}\left[ x \right]} \right) - xf\left( x \right)\frac{d}{{dx}}\left[ {g\left( x \right)} \right]}}{{{{\left[ {g\left( x \right)} \right]}^2}}} \cr
& {\text{solve the derivatives}} \cr
& \frac{d}{{dx}}\left[ {\frac{{xf\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{g\left( x \right)\left( {xf'\left( x \right) + f\left( x \right)\left( 1 \right)} \right) - xf\left( x \right)g'\left( x \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}} \cr
& {\text{simplifying}} \cr
& \frac{d}{{dx}}\left[ {\frac{{xf\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{xg\left( x \right)f'\left( x \right) + f\left( x \right)g\left( x \right) - xf\left( x \right)g'\left( x \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}} \cr} $$
