Answer
$\sqrt {7}\left( x+\dfrac {1}{x^{2}}\right) ^{\sqrt {7}-1}\left( 1-\dfrac {2}{x^{3}}\right) $
Work Step by Step
$y=\left( x+\dfrac {1}{x^{2}}\right) ^{\sqrt {7}}=y'=\sqrt {7}\times \left( x+x^{-2}\right) ^{\sqrt {7}-1}\times \left( \dfrac {d}{dx}\left( x+x^{-2}\right) \right) =\sqrt {7}\left( x+\dfrac {1}{x^{2}}\right) ^{\sqrt {7}-1}\left( 1-\dfrac {2}{x^{3}}\right) $
