(a) grad(phi) = <phi_x, phi_y, phi_z>, where phi_x is the partial derivative of phi with respect to x. Since phi_x=0, phi_y=sin(-z)*e^(-y)*(-1)=-sin(-z)*e^(-y), and phi_z=cos(-z)*(-1)*e^(-y)=-cos(-z)*e^(-y), then we have
grad(phi)=<0, -sin(-z)*e^(-y), -cos(-z)*e^(-y)>.
(b) The Laplacian of phi is L(phi)=phi_xx + phi_yy + phi_zz, where phi_xx is the second partial derivative of phi. Note that phi_xx=0, phi_yy=-sin(-z)*e^(-y)*(-1)=sin(-z)*e^(-y), and phi_zz=-(-sin(-z))*(-1)*e^(-y)=-sin(-z)*e^(-y). Therefore, L(phi)=0+sin(-z)*e^(-y) + (-sin(-z)*e^(-y)) =0.
(c) Since the Laplacian of phi is 0 by part (b), then we conclude that the scalar field is harmonic.