Answer
$$\nabla z = \left( {42/y} \right)\cos \left( {6x/y} \right)\left[ {{\bf{i}} + \left( { - x/y} \right){\bf{j}}} \right]$$
Work Step by Step
$$\eqalign{
& z = 7\sin \left( {6x/y} \right) \cr
& {\text{calculate the partial derivatives }}{z_x}{\text{ and }}{z_y} \cr
& {z_x} = \frac{\partial }{{\partial x}}\left[ {7\sin \left( {6x/y} \right)} \right] \cr
& {\text{treat }}y{\text{ as a constant}}{\text{; recall that }}\frac{d}{{dx}}\left[ {\sin u} \right] = \cos u\frac{{du}}{{dx}} \cr
& {z_x} = 7\cos \left( {6x/y} \right)\frac{\partial }{{\partial x}}\left[ {6x/y} \right] \cr
& {z_x} = 7\cos \left( {6x/y} \right)\left( {6/y} \right) \cr
& {z_x} = \left( {42/y} \right)\cos \left( {6x/y} \right) \cr
& and \cr
& {z_y} = \frac{\partial }{{\partial y}}\left[ {7\sin \left( {6x/y} \right)} \right] \cr
& {\text{treat }}y{\text{ as a constant}}{\text{; recall that }}\frac{d}{{dy}}\left[ {\sin u} \right] = \cos u\frac{{du}}{{dy}} \cr
& {z_y} = 7\cos \left( {6x/y} \right)\left( { - 6/{y^2}} \right) \cr
& {z_y} = \left( { - 42x/{y^2}} \right)\cos \left( {6x/y} \right)\left( { - 6x/{y^2}} \right) \cr
& \cr
& {\text{The gradient of the function }}z\left( {x,y} \right){\text{ is defined by }}\left( {{\text{see page 963}}} \right) \cr
& \nabla z = {z_x}{\bf{i}} + {z_y}{\bf{j}} \cr
& {\text{}}{\text{substituting the partial derivatives we obtain}} \cr
& \nabla z = \left( {42/y} \right)\cos \left( {6x/y} \right){\bf{i}} + \left( { - 42x/{y^2}} \right)\cos \left( {6x/y} \right){\bf{j}} \cr
& {\text{factoring}} \cr
& \nabla z = \left( {42/y} \right)\cos \left( {6x/y} \right)\left[ {{\bf{i}} + \left( { - x/y} \right){\bf{j}}} \right] \cr} $$
