Answer
$$\nabla {\text{ }}f\left( {x,y} \right) = \cos \left( {3x + 2y} \right)\left\langle {3,2} \right\rangle {\text{ and }}\nabla {\text{ }}f\left( {\pi ,3\pi /2} \right) = \left\langle {3,2} \right\rangle $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \sin \left( {3x + 2y} \right);\,\,\,\,\,\,P\left( {\pi ,3\pi /2} \right) \cr
& {\text{compute the gradient of }}f\left( {x,y} \right),{\text{ }} \cr
& {\text{first find the partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right).{\text{ then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\sin \left( {3x + 2y} \right)} \right] \cr
& {\text{treat }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \cos \left( {3x + 2y} \right)\frac{\partial }{{\partial x}}\left[ {\left( {3x + 2y} \right)} \right] \cr
& {f_x}\left( {x,y} \right) = 3\cos \left( {3x + 2y} \right) \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\sin \left( {3x + 2y} \right)} \right] \cr
& {\text{treat }}x{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \cos \left( {3x + 2y} \right)\frac{\partial }{{\partial y}}\left[ {\left( {3x + 2y} \right)} \right] \cr
& {f_x}\left( {x,y} \right) = 2\cos \left( {3x + 2y} \right) \cr
& {\text{using }}\nabla {\text{ }}f\left( {x,y} \right) = {f_x}\left( {x,y} \right){\bf{i}} + {f_y}\left( {x,y} \right){\bf{j}}.{\text{ then substituting the partial}} \cr
& {\text{derivatives}} \cr
& \nabla {\text{ }}f\left( {x,y} \right) = 3\cos \left( {3x + 2y} \right){\bf{i}} + 2\cos \left( {3x + 2y} \right){\bf{j}} \cr
& or \cr
& \nabla {\text{ }}f\left( {x,y} \right) = \left\langle {3\cos \left( {3x + 2y} \right),2\cos \left( {3x + 2y} \right)} \right\rangle \cr
& \nabla {\text{ }}f\left( {x,y} \right) = \cos \left( {3x + 2y} \right)\left\langle {3,2} \right\rangle \cr
& {\text{evaluate }}\nabla {\text{ }}f\left( {x,y} \right){\text{ at the point }}P \cr
& \nabla {\text{ }}f\left( {\pi ,3\pi /2} \right) = \cos \left( {3\left( \pi \right) + 2\left( {\frac{{3\pi }}{2}} \right)} \right)\left\langle {3,2} \right\rangle \cr
& \nabla {\text{ }}f\left( {\pi ,3\pi /2} \right) = \cos \left( {6\pi } \right)\left\langle {3,2} \right\rangle \cr
& \nabla {\text{ }}f\left( {\pi ,3\pi /2} \right) = \left\langle {3,2} \right\rangle \cr
& \cr
& \nabla {\text{ }}f\left( {x,y} \right) = \cos \left( {3x + 2y} \right)\left\langle {3,2} \right\rangle {\text{ and }}\nabla {\text{ }}f\left( {\pi ,3\pi /2} \right) = \left\langle {3,2} \right\rangle \cr} $$