Answer
$$h'\left( x \right) = \frac{{{e^{2{x^2} - x + 1/x}}\left( {4{x^3} - {x^2} - 1} \right)}}{{{x^2}}}$$
Work Step by Step
$$\eqalign{
& h\left( x \right) = {e^{2{x^2} - x + 1/x}} \cr
& {\text{Differentiate}} \cr
& h'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{2{x^2} - x + 1/x}}} \right] \cr
& {\text{Use the rule }}\frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}} \cr
& h'\left( x \right) = {e^{2{x^2} - x + 1/x}}\frac{d}{{dx}}\left[ {2{x^2} - x + \frac{1}{x}} \right] \cr
& {\text{Compute derivatives and simplify, recall that }}\frac{d}{{dx}}\left[ {{x^{ - 1}}} \right] = - {x^{ - 2}} \cr
& h'\left( x \right) = {e^{2{x^2} - x + 1/x}}\left( {4x - 1 - {x^{ - 2}}} \right) \cr
& h'\left( x \right) = {e^{2{x^2} - x + 1/x}}\left( {4x - 1 - \frac{1}{{{x^2}}}} \right) \cr
& h'\left( x \right) = {e^{2{x^2} - x + 1/x}}\left( {\frac{{4{x^3} - {x^2} - 1}}{{{x^2}}}} \right) \cr
& h'\left( x \right) = \frac{{{e^{2{x^2} - x + 1/x}}\left( {4{x^3} - {x^2} - 1} \right)}}{{{x^2}}} \cr} $$
