= Solution
Consider the globally smooth potential $\Phi(x,y,z)=xy^2\sin(xz)$. Direct differentiation gives
$$
\Phi_x=y^2\sin(xz)+xy^2z\cos(xz)=u,\qquad
\Phi_y=2xy\sin(xz)=v,\qquad
\Phi_z=x^2y^2\cos(xz)=w.
$$
Thus the <differential one-form> is the <exact differential> $d\Phi$, proving exactness rather than only checking a necessary condition on its partial derivatives. Along any piecewise differentiable path, the <chain rule> gives $u\,dx+v\,dy+w\,dz=d\Phi$. The <fundamental theorem of calculus> therefore makes the <line integral> depend only on its endpoints:
$$
\boxed{\int_{(0,0,0)}^{(\pi/2,1,1)}(u\,dx+v\,dy+w\,dz)
=\Phi(\pi/2,1,1)-\Phi(0,0,0)=\frac\pi2.}
$$
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