Does the gradient of a gradient exist? Does the gradient of a gradient exist and if yes, what is it and does it have a intuitive representation ?
Thanks for any answers!
 A: The gradient of a gradient is commonly called the second derivative. A hill with a constant slope has a gradient, but the gradient of the gradient is zero. If the hill gets steeper as you get near the top then the  gradient of the gradient would be a positive number.
A: There is the array of gradients of each component of $\nabla f(x_1,\dots,x_n)$, that is called hessian matrix of $f$ (usually denoted by $\nabla^2 f$ in nonlinear optimization texts) or the jacobian of $\nabla f$. It is associated with the curvature of the function in each point.
A: Exactly what do you mean by "gradient"?  To those who speak "British English", the "gradient" of a function is just what those who speak "American English" would call the "derivative".  If that is what you mean then, yes, the "gradient of a gradient" is the "second derivative".  For example, if $f(x)= x^3$ then the gradient of f is the first derivative, $f'(x)= 3x^2$ and the "gradient of the gradient", or "second derivative" is the derivative of that: $f''(x)= 6x$.
But to those who speak "American English" and say "derivative" rather than "gradient", the gradient is a vector-valued operator.  The "gradient" is applied to a scalar valued function of several variables and results in a vector valued function.  Given a function of more than one variable, the gradient of that function is the vector, each of whose components is the derivative in that direction.  If $f(x, y, z)= x^2ye^z$ then the "gradient" of f is $\nabla f= 2xye^z\vec{i}+ x^2e^z\vec{j}+ x^2e^z\vec{k}$.  Since the "gradient", in this sense, must be applied to a scalar valued function and the result is a vector valued function, no you cannot apply the gradient a second time.  There is no "gradient of the gradient". 
A: In addition, the "nabla", $\nabla$, can be used as the "divergence" or the "curl" both of which apply to vectors.  With $\vec{f}= f_x\vec{i}+ f_y\vec{j}+ f_z\vec{k}$,  $div \vec{f}= \nabla \cdot \vec{x}(x, y, z)= \frac{\partial f_x}{\partial x}+ \frac{\partial f_y}{\partial y}+ \frac{\partial f_z}{\partial z}$ and $curl \vec{f}= \nabla\times \vec{f}= \left(\frac{\partial f_y}{\partial z}- \frac{\partial f_z}{\partial y}\right)\vec{i}+ \left(\frac{\partial f_z}{\partial x}- \frac{\partial f_x}{\partial z}\right)\vec{j}+ \left(\frac{\partial f_x}{\partial y}- \frac{\partial f_y}{\partial x}\right)\vec{k}$.  
Because these can be applied to vector valued functions, we can apply either of them to the vector valued "grad f".  In particular, $\nabla\cdot \nabla f$ takes scalar valued f to a scalar valued function.  That is often symbolized as $\nabla^2 f$ or (more physics texts) $\triangle f.  As long as the function has differentiable partials so that the mixed derivatives are equal, "curl grad f" is the 0 vector so you never see that. 
