The problem i'm trying to solve is
$(\mathbf{n \times \bigtriangledown \times n})^2$
$\mathbf{n \times \bigtriangledown \times n = n_{30}}$
where $\mathbf{n}=<\cos(\theta)\cos(\phi),\cos(\theta)\sin(\phi),\sin(\theta)>$, $\bigtriangledown=\frac{\partial}{\partial x}+\frac{\partial}{\partial y}+\frac{\partial}{\partial z}$ and $\theta=\theta(z) ,\phi=\phi(z)$ i.e. $\frac{\partial \mathbf{n}}{\partial x}=\frac{\partial \mathbf{n}}{\partial y}=0 $
The answer i have right now is
$\mathbf{n_{30}}=[\sin^2(\theta)\cos(\phi)(\frac{d\theta}{dz})+\sin(\theta)\cos(\theta)\sin(\phi)(\frac{d\phi}{dz})]\mathbf{i}+[\sin^2(\theta)\sin(\phi)(\frac{d\theta}{dz})-\sin(\theta)\cos(\theta)\cos(\phi)(\frac{d\phi}{dz})]\mathbf{j}+[-\cos(\theta)\sin(\theta)(\frac{d\theta}{dz})+\cos^2(\theta)(\cos^2(\phi)+\cos(\phi)\sin(\phi))(\frac{d\phi}{dz})]\mathbf{k}$
As far as the main problem, i'm not sure if it's correct or not because of how many terms i have.. but i know i'm supposed to get something along the lines of $(\mathbf{n \times \bigtriangledown \times n})^2=\sin^2(\theta)(\frac{d\theta}{dz})^2+\sin^2(\theta)\cos^2(\theta)(\frac{d\phi}{dz})^2$
Additional details:
For now, i am not sure how to expand $\mathbf{n \times \bigtriangledown \times n}$, i posted a question earlier about how using the relation $A \times B \times C= (A \cdot C)B-(A \cdot B)C$ gives $\mathbf{n \times \bigtriangledown \times n = (n \cdot n)\bigtriangledown - (n \cdot \bigtriangledown )n}$, i know this equation gives me a useful scalar, so $\bigtriangledown$ has to act on something to evaluate and not stay in front, similarly for expansion of $\mathbf{(n \times \bigtriangledown \times n)^2}$..
If you keep an eye out for 'Pythagorean identities' $\sin^2(\alpha)+\cos^2(\alpha) = 1$ then the process simplifies quite a bit. For instance, looking at your $\mathbb{i}$ and $\mathbb{j}$ terms for $\mathbb{n}$, you'll notice that they're of the form $n_i = A\cos(\phi)+B\sin(\phi)$, $n_j=A\sin(\phi)-B\cos(\phi)$ for specific expressions $A$ and $B$; then $(n_i)^2=A^2\cos^2(\phi)+2AB\sin(\phi)\cos(\phi)+B^2\sin^2(\phi)$ and $(n_j)^2 = A^2\sin^2(\phi)-2AB\sin(\phi)\cos(\phi)+B^2\cos^2(\phi)$, so their sum is $(n_i)^2+(n_j)^2$ $=A^2\left(\cos^2(\phi)+\sin^2(\phi)\right)+B^2\left(\sin^2(\phi)+\cos^2(\phi)\right)$ $=A^2+B^2$. (You can also see this in 'coordinate-independent' fashion by recognizing the two-vector of the $\mathbb{ij}$ components of $\mathbb{n}$ as being the vector $(A,B)$ rotated by a quantity related to $\phi$; that 'sin/cos, -cos/sin' or more frequently 'cos/sin, -sin/cos' pattern is the harbinger of a 2d rotation.)