1. **State the problem:** We need to find the work done by the gradient field $\vec{F} = \langle yz, xz, xy \rangle$ when moving from point $A = (3,1,4)$ to point $B = (1,5,9)$.\n\n2. **Recall the formula:** For a gradient field $\vec{F} = \nabla f$, the work done moving from $A$ to $B$ is given by the difference in the potential function $f$ evaluated at $B$ and $A$: $$W = f(B) - f(A)$$\n\n3. **Find the potential function $f$:** Since $\vec{F} = \langle yz, xz, xy \rangle$, we have \begin{align*} \frac{\partial f}{\partial x} &= yz \\ \frac{\partial f}{\partial y} &= xz \\ \frac{\partial f}{\partial z} &= xy \end{align*} Integrate $\frac{\partial f}{\partial x} = yz$ with respect to $x$: $$f = xyz + g(y,z)$$ where $g(y,z)$ is an arbitrary function of $y$ and $z$.\n\n4. **Use $\frac{\partial f}{\partial y} = xz$ to find $g(y,z)$:** Differentiate $f$ with respect to $y$: $$\frac{\partial f}{\partial y} = xz + \frac{\partial g}{\partial y}$$ Set equal to $xz$: $$xz + \frac{\partial g}{\partial y} = xz \implies \frac{\partial g}{\partial y} = 0$$ So $g$ does not depend on $y$, $g = h(z)$.\n\n5. **Use $\frac{\partial f}{\partial z} = xy$ to find $h(z)$:** Differentiate $f$ with respect to $z$: $$\frac{\partial f}{\partial z} = xy + h'(z)$$ Set equal to $xy$: $$xy + h'(z) = xy \implies h'(z) = 0$$ So $h(z)$ is a constant, say $C$.\n\n6. **Potential function:** $$f(x,y,z) = xyz + C$$ We can ignore the constant $C$ since it cancels in the difference.\n\n7. **Calculate work done:** $$W = f(1,5,9) - f(3,1,4) = (1)(5)(9) - (3)(1)(4) = 45 - 12 = 33$$\n\n**Final answer:** The work done by the gradient field from $A$ to $B$ is $33$.