Differentiation in 3d of sin root and fractions in one! -> to find the normal to a function I have to find the normal of this function at a defined point $x$ and $z$, I have done A Level maths but that was some time ago but I don't think it was covered to this level, I am now doing a CS degree. I thought the best way would be to differentiate the function to get the tangent, then get the normal.
I have no idea where to start.
Any ideas on how I should go about it?  
$$y = \frac{\sin (\sqrt{x^2+z^2})}{\sqrt{x^2+z^2}}$$

The normal to a curve at a point is perpendicular to the gradient at that point.  In your case:
$$f(x,z) = \frac{\sin{\sqrt{x^2+z^2}}}{\sqrt{x^2+z^2}}$$
$$\begin{align}\nabla f &= \left ( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial z}\right ) \\ &= \left ( \frac{x \cos
   \left(\sqrt{x^2+z^2}\right)}{x^2+z^2
   }-\frac{x \sin
   \left(\sqrt{x^2+z^2}\right)}{\left(x
   ^2+z^2\right)^{3/2}}, \frac{z \cos
   \left(\sqrt{x^2+z^2}\right)}{x^2+z^2
   }-\frac{z \sin
   \left(\sqrt{x^2+z^2}\right)}{\left(x
   ^2+z^2\right)^{3/2}}\right )  \end{align}$$
The unit normal to the curve is then
$$\frac{\left ( -\frac{\partial f}{\partial z}, \frac{\partial f}{\partial x}\right )}{\sqrt{\left ( \frac{\partial f}{\partial x} \right )^2+\left ( \frac{\partial f}{\partial z} \right )^2}} $$
