Answer
$$\frac{{1/2}}{{x - 4}} + \frac{{1/2}}{{x + 4}}$$
Work Step by Step
$$\eqalign{
& \frac{{{x^2}}}{{{x^3} - 16x}} \cr
& = \frac{{{x^2}}}{{x\left( {{x^2} - 16} \right)}} = \frac{x}{{{x^2} - 16}} \cr
& {\text{factoring}} \cr
& = \frac{x}{{\left( {x - 4} \right)\left( {x + 4} \right)}} \cr
& {\text{partial fraction decomposition}} \cr
& \frac{x}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = \frac{A}{{x - 4}} + \frac{B}{{x + 4}} \cr
& x = A\left( {x + 4} \right) + B\left( {x - 4} \right) \cr
& {\text{for }}x = 4 \cr
& 4 = A\left( {4 + 4} \right) + B\left( {4 - 4} \right) \cr
& 4 = 8A \cr
& 1/2 = A \cr
& {\text{for }}x = - 4 \cr
& - 4 = A\left( { - 4 + 4} \right) + B\left( { - 4 - 4} \right) \cr
& - 4 = - 8B \cr
& 1/2 = B \cr
& {\text{substituting the values}} \cr
& \frac{A}{{x - 4}} + \frac{B}{{x + 4}} = \frac{{1/2}}{{x - 4}} + \frac{{1/2}}{{x + 4}} \cr} $$
