Let
.
I don't see how $\frac{\partial g}{\partial y} = 0$. I don't even know what g(y,z) is supposed to be. We haven't defined the function g(y,z). Is g(y,z) supposed to be the integral of $N$ with respect to $ y$ plus the integral of $ P$ with respect to $ z$ ?
How do we prove that g(y,z) is a function of z alone?
Why did we integrate M with respect to x rather than integrate N with respect to y or P with respect to z?
We did integration on the variable $ x$ to derive $ f(x,y)= "e^xcos(y)+xyz"$. Why aren't we using integration on the variable $ y$ of the function $N$ or $P$ to derive the potential function?
I'm in disbelief that $ h(z)=z^2/2+C$ I think that $ h(z)$ should be equal to the integral of $ P$ with respect to $ z$ or $ xyz+z^2$.
First we check that $\nabla\times{\bf F} = {\bf 0}$. Now we know that a potential, $f(x,y,z)$ such that ${\bf F} = \nabla f$, exists. Thus, $$(e^x\cos y+yz,xz-e^x\sin y,xy+z) = (f_x,f_y,f_z)$$ for some $f(x,y,z)$. It must be the case that $f_x = e^x\cos y +yz$ and so by integrating with respect to $x$ we find $$f(x,y,z) = e^x\cos y + xyz + g(y,z).$$ The function $g(y,z)$ is an integration constant. (Note that $g_x = 0$, that is, $g(y,z)$ is a constant with respect to $x$.) We must also have $$f_y = xz-e^x\sin y.$$ But $f_y = -e^x\sin y + xz + g_y$. Thus, $g_y = 0$ and so $g(y,z) = h(z)$. Lastly, $$f_z = xy+z.$$ But $f_z = xy + h_z$ and so $h_z = z$. Thus, $h(z) = z^2/2 + C$, where $C$ is an arbitrary constant.