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The potential of F is any function f such that del f = F.

I got this:
xycos(z), xycos(z), -xysin(z).
Is this correct?

There is a methodical way of doing this.

I did the integration part and got the above answer but how do I find the constant of integration?

\[
\frac{\partial}{\partial x}f(x,y,z)=y\sin(z) \implies f(x, y, z) = xy\sin(z)+g(y, z)
\]

So then you integrate another function to solve for \(g(y, z)\).

From what I'm seeing, wio's way of doing this will get you to a solution, rather easily.

It seems that \(g'(y, z) = 0\). So we know that \(g(y, z)\) is a constant with respect to \(y\).

wonder why none of the local residents just give you the answer

^^because giving answers for free is against the Code of Conduct.

what is "giving the answer for free?"

^^well said.