The curl is a little more work but still just formula work so here is the curl.

\[\begin{align*}{\mathop{\rm curl}\nolimits} \vec F& = \nabla \times \vec F = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{\displaystyle \frac{\partial }{{\partial x}}}&{\displaystyle \frac{\partial }{{\partial y}}}&{\displaystyle \frac{\partial }{{\partial z}}}\\{3x + 2{z^2}}&{\displaystyle \frac{{{x^3}{y^2}}}{z}}&{7x - z}\end{array}} \right|\\ & = \frac{\partial }{{\partial y}}\left( {7x - z} \right)\vec i + \frac{\partial }{{\partial z}}\left( {3x + 2{z^2}} \right)\vec j + \frac{\partial }{{\partial x}}\left( {\frac{{{x^3}{y^2}}}{z}} \right)\vec k\\ & \hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \frac{\partial }{{\partial y}}\left( {3x + 2{z^2}} \right)\vec k - \frac{\partial }{{\partial x}}\left( {7x - z} \right)\vec j - \frac{\partial }{{\partial z}}\left( {\frac{{{x^3}{y^2}}}{z}} \right)\vec i\\ & = 4z\vec j + \frac{{3{x^2}{y^2}}}{z}\vec k - 7\vec j + \frac{{{x^3}{y^2}}}{{{z^2}}}\vec i\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{{x^3}{y^2}}}{{{z^2}}}\vec i + \left( {4z - 7} \right)\vec j + \frac{{3{x^2}{y^2}}}{z}\vec k}}\end{align*}\]

Again, don’t forget the minus sign on the 3^rd component.