Finding a vector that has 0 curl and 0 div SO as stated, I am trying to find a vector $\vec F$such that
$$\nabla \times \vec F=0$$
$$\nabla \cdot \vec F=0$$
The way I go about it is:
Becasue curl is 0, we know that $\vec F=\nabla f$ so the divergence equation then becomes
$$\nabla ^2f=0$$
Which I then say $f=A(x)B(y)C(z)$, which results in me getting the following function f:
$$f=\cos(x)\cosh(y)\cos(z)$$
so $\vec F=\nabla f$, 
Which has 0 curl, but nonzero div. Sad face
 A: There are many ways to generate harmonic functions. 
One way is as the poster suspected by looking at functions of the form:
$$f(x) = \cos(p x) \cosh(q y) \cos( r z)$$
By direct substitution, it is easy to see
$$\nabla^2 f(x) = (q^2 - p^2 - r^2) f(x)$$
If one has choose $p, q, r$ such that $q^2 - p^2 - r^2 = 0$,  e.g. 
$$p = r = 1 \quad\text{ and }\quad q = \sqrt{2},$$ 
then we get a $f$ that is harmonic, i.e. $\displaystyle\nabla^2 f = 0$. 
We can also generate polynomials that is harmonic. The simplest way is to
observe:
$$\begin{cases}
\frac{\partial^2}{\partial x^2} (x \pm iy)^n = \;\;n(n-1)(x \pm iy)^{n-2}\\
\frac{\partial^2}{\partial y^2} (x\pm iy)^n = -n(n-1)(x\pm iy)^{n-2}
\end{cases}$$
This implies $\displaystyle\quad\nabla^2 ( x \pm i y)^n = 0\quad$ and hence
$$\phi(x) = (x + iy)^n + (x - iy)^n$$ 
is a polynomial that is harmonic. By a orthogonal transformation of $(x,y,z)$ and through linear combinations, you can generate other polynomials of degree $n$ that is harmonic and homogenous in $(x,y,z)$. 
A: i remember thinking some where that all harmonic functions have zero curl 
more precisely all conservative vector fields are harmonic also follow Clairaut's Theorem 
which is second partials in reverse order are equal 
and for divergence to e zero the vector field must be constant meaning no rate of change in all directions 
an example in 3D 
f=xyz
F= yz i + xz j + xy K 
Curl = 0 ( check this by computing but it is 0) 
Div = 0
so F sub xy = F sub yx so they are conservative so no curl and clearly no divergence 
A: yankeefan11, you were pretty close!  Using the bold face font for vectors, we have as a given that
$\nabla  \times \mathbf{F} = 0, \tag{1}$
which does in fact imply the existence of $\phi$ such that
$\mathbf F = \nabla \phi, \tag{2}$
and now using the divergence condition we find we must have
$\nabla^2 \phi = \nabla \cdot \mathbf F = 0. \tag{3}$
Running things in reverse, we see that for any harmonic $\phi$, setting 
$\mathbf F = \nabla \phi \tag{4}$
produces an $\mathbf F$ with 
$\nabla \cdot \mathbf F = 0 \tag{5}$
and
$\nabla \times \mathbf F = 0 \tag{6}$
as well.  There are lots of solutions; the field gets considerably thinner when boundary conditions are added.
Fiat Lux!
A: A simpler example:
Consider $F(x,yz) = (-y/(x²+y²),x/(x²+y²),0)$
The jacobian matrix $DF$ shows that $tr(DF)$ = div $F = 0$, and since it is symmetric, it follows that curl $F$ = 0.  
