Curl of gradient of potential in electrostatic We know, curl of E is zero (this field is conservative). Again 
E=-grad V. So, we get curl of (-grad V)=0, i.e. curl of gradient of potential is zero. Is there any condition on potential?
 A: Try to manually compute (i.e. write it out in it's explicit form) the gradient of some potential $V$, and then compute its curl. 
This is more than sufficient to find out if some condition is needed on $V$.
(If you are stuck, check out the theorems on: https://en.wikipedia.org/wiki/Symmetry_of_second_derivatives)
A: As hrithik says curl of a gradient of  is always zero. Let V=V(x, y, z).  The gradient of V  ie $\nabla V= \frac{\partial V} {\partial x} \hat{i} +\frac{\partial V} {\partial y} \hat{j} +\frac{\partial V} {\partial z} \hat{k} $. Now the curl of grad v is ie $\nabla × \nabla V$ now you got a determinate. Upon solving that $\frac{\partial} {\partial y} ({\frac{\partial V} {\partial z}) \hat{i}} - \frac{\partial} {\partial z} ({\frac{\partial V} {\partial y}) \hat{i}} - \frac{\partial} {\partial x} ({\frac{\partial V} {\partial z}) \hat{j}} + \frac{\partial} {\partial z} ({\frac{\partial V} {\partial x}) \hat{j}} + \frac{\partial} {\partial x} ({\frac{\partial V} {\partial y}) \hat{k}-\frac{\partial} {\partial y} ({\frac{\partial V} {\partial x}) \hat{k}}}$ now basically $\frac{\partial} {\partial y} ({\frac{\partial V} {\partial z}) } = \frac{\partial} {\partial z} ({\frac{\partial V} {\partial y}) }$ as well $\frac{\partial} {\partial x} ({\frac{\partial V} {\partial z})}$ = $ \frac{\partial} {\partial z} ({\frac{\partial V} {\partial x})} $ and $\frac{\partial} {\partial x} ({\frac{\partial V} {\partial y})} = \frac{\partial} {\partial y} ({\frac{\partial V} {\partial x})} $ so curl of gradient of V or any other function is always zero.  So no charactesic of V is needed except for it should exist ie the electric field should be Conservative. 
