Answer
$\vec{v} = (yz)\hat{i}+ (xz)\hat{j}+ (xy)\hat{k}$ has a divergence and curl of zero.
Work Step by Step
Let's assume that $\vec{v} = \nabla f$ for some potential function $f$
For the divergence to be zero
$\frac{\partial v_x}{\partial x} = 0, \frac{\partial v_y}{\partial y} = 0 , \frac{\partial v_z}{\partial z} = 0$
This will always be true if $v_x,v_y,v_z$ do not explicitly depend on on $x,y,z$ respectively
So $\nabla f = f(y,z) \hat{i} + f(x,z) \hat{j} + f(x,y)\hat{k}$
$f(x,y,z) = xyz$ satisfies this condition
$\nabla f = yz \hat{i} + xz \hat{j} + xy \hat{k}$
$\nabla \cdot (yz \hat{i} + xz \hat{j} + xy \hat{k}) = 0+0+0 = 0$
Let's see if this satisfies $\nabla \times \vec{v} = 0$
$Curl_i = \frac{\partial (xy)}{\partial y} - \frac{\partial (xz)}{\partial z} = 0$
$Curl_j = -\frac{\partial (xy)}{\partial x} + \frac{\partial (yz)}{\partial z} = 0$
$Curl_k = \frac{\partial (xz)}{\partial x} - \frac{\partial (yz)}{\partial y} = 0$
$\nabla \times \vec{v} = 0$
$\vec{v} = yz \hat{i} + xz \hat{j} + xy \hat{k}$ satisfies both conditions
