Showing that a vector field is conservative without finding a potential function

Question

I am really new to this topic and was tasked to show that $$G (x,y,z)=\langle z^3−3y^2e^{3x},−2ye^{3x}+2\cos z,4+3xz^2−2y\sin z\rangle$$ is conservative without finding a potential function. How do I pull this off? I have no idea where to start with this one. Any help would be much appreciated.

Answer 1

Let $M = z^3-3y^2e^{3x}, N = -2ye^{3x}+2\cos z, P = 4+3xz^2 - 2y\sin z$. Then $\dfrac{\partial M}{\partial y}= \dfrac{\partial N}{\partial x}= -6ye^{3x}, \dfrac{\partial M}{\partial z} = \dfrac{\partial P}{\partial x} = 3z^2, \dfrac{\partial N}{\partial z} = \dfrac{\partial P}{\partial y} = -2\sin z\implies \text{Curl}(G) = 0\implies G $ is conservative.