Answer
$${\text{No horizontal tangent lines}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{{x^2}}}{2} - \frac{8}{{{x^2}}} \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{2} - \frac{8}{{{x^2}}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{2}} \right] - \frac{d}{{dx}}\left[ {\frac{8}{{{x^2}}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\frac{d}{{dx}}\left[ {{x^2}} \right] - 8\frac{d}{{dx}}\left[ {{x^{ - 2}}} \right] \cr
& {\text{Apply }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}\left( {2x} \right) - 8\left( { - 2{x^{ - 3}}} \right) \cr
& \frac{{dy}}{{dx}} = x + 16{x^{ - 3}} \cr
& {\text{The tangent line to the graph is horizontal when }}\frac{{dy}}{{dx}} = 0,{\text{ then}} \cr
& x + 16{x^{ - 3}} = 0 \cr
& {\text{Solve for }}x \cr
& x + \frac{{16}}{{{x^3}}} = 0 \cr
& x = - \frac{{16}}{{{x^3}}} \cr
& {x^4} = - 16 \cr
& x = \pm \root 4 \of { - 16} \cr
& {\text{No real solutions, then there are no points at which the graph}} \cr
& {\text{has horizontal tangent lines}}{\text{.}} \cr} $$
