Answer
$$\frac{{\partial z}}{{\partial x}} = - \frac{{{z^2}}}{{2xz - y\cos z}}{\text{ and }}\frac{{\partial z}}{{\partial x}} = \frac{{\sin z}}{{2xz - y\cos z}}$$
Work Step by Step
$$\eqalign{
& x{z^2} - y\sin z = 0 \cr
& {\text{Letting }}F\left( {x,y,z} \right) = x{z^2} - y\sin z \cr
& {\text{Calculate }}{F_x}\left( {x,y,z} \right),{\text{ }}{F_y}\left( {x,y,z} \right){\text{ and }}{F_z}\left( {x,y,z} \right) \cr
& {F_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {x{z^2} - y\sin z} \right] = {z^2} \cr
& {F_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {x{z^2} - y\sin z} \right] = - \sin z \cr
& {F_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {x{z^2} - y\sin z} \right] = 2xz - y\cos z \cr
& {\text{Using Theorem 13}}.{\text{8}},{\text{ you have}} \cr
& \frac{{\partial z}}{{\partial x}} = - \frac{{{F_x}\left( {x,y,z} \right)}}{{{F_z}\left( {x,y,z} \right)}} = - \frac{{{z^2}}}{{2xz - y\cos z}} \cr
& \frac{{\partial z}}{{\partial y}} = - \frac{{{F_y}\left( {x,y,z} \right)}}{{{F_z}\left( {x,y,z} \right)}} = \frac{{\sin z}}{{2xz - y\cos z}} \cr
& \cr
& \frac{{\partial z}}{{\partial x}} = - \frac{{{z^2}}}{{2xz - y\cos z}}{\text{ and }}\frac{{\partial z}}{{\partial x}} = \frac{{\sin z}}{{2xz - y\cos z}} \cr} $$
