Answer
$${s_y}\left( {y,z} \right) = {z^3}{\sec ^2}yz,\,\,\,\,\,\,\,\,\,{s_z}\left( {y,z} \right) = y{z^2}{\sec ^2}yz + 2z\tan yz$$
Work Step by Step
$$\eqalign{
& s\left( {y,z} \right) = {z^2}\tan yz \cr
& {\text{Find the partial derivatives }}{s_y}\left( {y,z} \right){\text{ and }}{s_z}\left( {y,z} \right){\text{ then}} \cr
& {s_y}\left( {y,z} \right) = \frac{\partial }{{\partial y}}\left[ {{z^2}\tan yz} \right] \cr
& {\text{treat }}z{\text{ as a constant}} \cr
& {s_y}\left( {y,z} \right) = {z^2}\frac{\partial }{{\partial y}}\left[ {\tan yz} \right] \cr
& {s_y}\left( {y,z} \right) = {z^2}{\sec ^2}yz\frac{\partial }{{\partial y}}\left( {yz} \right) \cr
& {s_y}\left( {y,z} \right) = {z^2}{\sec ^2}yz\left( z \right) \cr
& {s_y}\left( {y,z} \right) = {z^3}{\sec ^2}yz \cr
& and \cr
& {s_z}\left( {y,z} \right) = \frac{\partial }{{\partial z}}\left[ {{z^2}\tan yz} \right] \cr
& {\text{treat }}y{\text{ as a constant}} \cr
& {s_z}\left( {y,z} \right) = {z^2}\frac{\partial }{{\partial z}}\left[ {\tan yz} \right] + \tan yz\frac{\partial }{{\partial z}}\left[ {{z^2}} \right] \cr
& {s_z}\left( {y,z} \right) = {z^2}\left( {y{{\sec }^2}yz} \right) + \tan yz\left( {2z} \right) \cr
& {s_z}\left( {y,z} \right) = y{z^2}{\sec ^2}yz + 2z\tan yz \cr} $$
