We know all we need to do here is compute the curl of the vector field.

\[\begin{align*}{\mathop{\rm curl}\nolimits} \vec F = \nabla \times \vec F &= \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{\displaystyle \frac{\partial }{{\partial x}}}&{\displaystyle \frac{\partial }{{\partial y}}}&{\displaystyle \frac{\partial }{{\partial z}}}\\{\displaystyle 4{y^2} + \frac{{3{x^2}y}}{{{z^2}}}}&{\displaystyle 8xy + \frac{{{x^3}}}{{{z^2}}}}&{\displaystyle 11 - \frac{{2{x^3}y}}{{{z^3}}}}\end{array}} \right|\\ & = \frac{\partial }{{\partial y}}\left( {11 - \frac{{2{x^3}y}}{{{z^3}}}} \right)\vec i + \frac{\partial }{{\partial z}}\left( {4{y^2} + \frac{{3{x^2}y}}{{{z^2}}}} \right)\vec j + \frac{\partial }{{\partial x}}\left( {8xy + \frac{{{x^3}}}{{{z^2}}}} \right)\vec k\\ & \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\,\,\,\,\,\, - \frac{\partial }{{\partial y}}\left( {4{y^2} + \frac{{3{x^2}y}}{{{z^2}}}} \right)\vec k - \frac{\partial }{{\partial x}}\left( {11 - \frac{{2{x^3}y}}{{{z^3}}}} \right)\vec j - \frac{\partial }{{\partial z}}\left( {8xy + \frac{{{x^3}}}{{{z^2}}}} \right)\vec i\\ & = - \frac{{2{x^3}}}{{{z^3}}}\vec i - \frac{{6{x^2}y}}{{{z^3}}}\vec j + \left( {8y + \frac{{3{x^2}}}{{{z^2}}}} \right)\vec k - \left( {8y + \frac{{3{x^2}}}{{{z^2}}}} \right)\vec k + \frac{{6{x^2}y}}{{{z^3}}}\vec j + \frac{{2{x^3}}}{{{z^3}}}\vec i\\ & = \underline {\vec 0} \end{align*}\] Show Step 2