Answer
$$\nabla z = - 6\sin \left( {25} \right){\bf{i}} - 8\sin \left( {25} \right){\bf{j}}$$
Work Step by Step
$$\eqalign{
& z = \cos \left( {{x^2} + {y^2}} \right),{\text{ }}\left( {3, - 4} \right) \cr
& {\text{Calculate the partial derivatives }}{z_x}{\text{ and }}{z_y} \cr
& {\text{ }}{z_x} = \frac{\partial }{{\partial x}}\left[ {\cos \left( {{x^2} + {y^2}} \right)} \right] \cr
& {\text{ }}{z_x} = - \sin \left( {{x^2} + {y^2}} \right)\frac{\partial }{{\partial x}}\left[ {{x^2} + {y^2}} \right] \cr
& {\text{ }}{z_x} = - \sin \left( {{x^2} + {y^2}} \right)\left( {2x} \right) \cr
& {\text{ }}{z_x} = - 2x\sin \left( {{x^2} + {y^2}} \right) \cr
& and \cr
& {z_y} = \frac{\partial }{{\partial y}}\left[ {\cos \left( {{x^2} + {y^2}} \right)} \right] \cr
& {\text{ }}{z_y} = - \sin \left( {{x^2} + {y^2}} \right)\frac{\partial }{{\partial y}}\left[ {{x^2} + {y^2}} \right] \cr
& {\text{ }}{z_y} = - \sin \left( {{x^2} + {y^2}} \right)\left( {2y} \right) \cr
& {\text{ }}{z_y} = - 2y\sin \left( {{x^2} + {y^2}} \right) \cr
& \cr
& {\text{The gradient of }}z{\text{ is}} \cr
& \nabla z = {z_x}{\bf{i}} + {z_y}{\bf{j}} \cr
& \nabla z = - 2x\sin \left( {{x^2} + {y^2}} \right){\bf{i}} - 2y\sin \left( {{x^2} + {y^2}} \right){\bf{j}} \cr
& {\text{At the point }}\left( {3, - 4} \right){\text{ the gradient is}} \cr
& \nabla z = - 2\left( 3 \right)\sin \left( {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2}} \right){\bf{i}} - 2\left( 4 \right)\sin \left( {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2}} \right){\bf{j}} \cr
& \nabla z = - 6\sin \left( {25} \right){\bf{i}} - 8\sin \left( {25} \right){\bf{j}} \cr} $$