Answer
$\text{Potential function: } f(x, y, z) = \frac{xz}{y} + C$
Work Step by Step
\[
\mathbf{F}(x, y, z) = \frac{z}{y}\mathbf{i} - \frac{xz}{y^2}\mathbf{j} + \frac{x}{y}\mathbf{k}
\]
\[
\nabla \times \mathbf{F} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\
\dfrac{z}{y} & -\dfrac{xz}{y^2} & \dfrac{x}{y}
\end{vmatrix}
\]
\[
= \left(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\right)\mathbf{i}
- \left(\frac{\partial R}{\partial x} - \frac{\partial P}{\partial z}\right)\mathbf{j}
+ \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\mathbf{k}
\]
\[
= \left(-\frac{x}{y^2} - \left(-\frac{x}{y^2}\right)\right)\mathbf{i}
- \left(\frac{1}{y} - \frac{1}{y}\right)\mathbf{j}
+ \left(-\frac{z}{y^2} - \left(-\frac{z}{y^2}\right)\right)\mathbf{k}
\]
\[
= 0
\]
\[
\therefore \mathbf{F} \text{ is conservative.}
\]
\[
f(x, y, z) = \int f_x\,dx = \int \frac{z}{y}\,dx = \frac{xz}{y} + g(y, z)+K1
\]
\[
f(x, y, z) = \int f_y\,dy = \int -\frac{xz}{y^2}\,dy = \frac{xz}{y} + h(x, z)+K2
\]
\[
f(x, y, z) = \int f_z\,dz = \int \frac{x}{y}\,dz = \frac{xz}{y} + k(x, y) +K3
\]
\[
\boxed{f(x, y, z) = \frac{xz}{y} + K}
\]
