Solution: We recall that a vector field in $\mathbb{R} ^3$ is irrotational if its curl is zero. We also recall that for a vector field $\mathbf{v} = v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k}$, the curl is given by \[ \mathrm{curl}(\mathbf{v} ) = \nabla \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ v_1 & v_2 & v_3 \end{vmatrix}. \] Thus the curl of the given vector field will be \begin{align*} \nabla \times \mathbf{v} & = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ (y+z) & (z+x) & (x + y) \\ \end{vmatrix} \\ & = \left[ 1 - 1 \right] \mathbf{i} - \left[ 1 - 1 \right] \mathbf{j} + \left[ 1 -1 \right] \mathbf{k} = \mathbf{0} . \end{align*} Thus, the given vector filed id irrotational.

Since $\mathbf{v} $ is irrotational, we can find a scalar function (potential function) $\phi $ such that $\mathbf{v} = \mathrm{grad} (\phi)$. Thus we have \begin{equation}\label{eq:22Mar2025-1} (y + z)\mathbf{i} + (z + x) \mathbf{j} + (x + y)\mathbf{k} = \frac{\partial \phi }{\partial x}\mathbf{i} + \frac{\partial \phi }{\partial y} \mathbf{j} + \frac{\partial \phi }{\partial z} \mathbf{k}. \end{equation} This gives a system of equations \begin{align*} \frac{\partial \phi }{\partial x} = ( y + z) \implies \phi (x,y,z) & = \int (y +z) \ \mathrm{d} x + f(y,z) \\ & = x(y+z) + f(y,z), \end{align*} where $f(y,z)$ is a function of only $y$ and $z$.

Now differentiating $\phi $ with respect $y$ and from the equation \eqref{eq:22Mar2025-1}, \begin{align*} \frac{\partial \phi }{\partial y} = (z + x) & \implies x + \frac{\partial f}{\partial y} = z + x \\ & \implies \frac{\partial f}{\partial y} = z \\ & \implies f(y,z) = yz + g(z), \end{align*} where $g$ is the function of $z$ only. Thus, $\phi (x,y,z) = x(y+z) + yz + g(z)$. Finally, \begin{align*} \frac{\partial \phi }{\partial z} = (x + y) & \implies x + y + g'(z) = x + y \\ & \implies g'(z) = 0\\ & \implies g(z) = c \text{ (constant) }. \end{align*} Therefore, \[ \textcolor{blue}{ \boxed{ \phi (x,y,z) = xy + yz + zx + c } } \]