The function is:
$$f = \frac{\mathrm{cos}(\vec{a}\cdot \vec{r})}{r^2}$$
so I did the following:
$$(\vec{a},\vec{r}) = (xa_x, ya_y, za_z)$$ $$r = \sqrt{x^2+y^2+z^2}$$ $$ r^2 = x^2+y^2+z^2$$
and ended up with
$$ f =\frac{\mathrm{cos}(xa_x,ya_y,za_z)}{x^2+y^2+z^2}$$
what would the derivative be then?
Using the product rule $\nabla (fg)=f\nabla (g)+g\nabla(f)$ and the chain rule $\nabla f(g(\vec r))=f'(g(\vec r))\nabla g(\vec r)$, we can write
$$\begin{align} \nabla f&=\nabla \left(\frac{\cos(\vec a\cdot \vec r)}{r^2}\right)\\\\ &=\frac1{r^2}\nabla (\cos(\vec a\cdot \vec r))+\cos(\vec a\cdot \vec r)\nabla \left(\frac{1}{r^2}\right)\\\\ &=-\frac{\sin(\vec a\cdot \vec r)}{r^2}\nabla (\vec a\cdot \vec r)-\frac{2\cos(\vec a\cdot \vec r)}{r^3}\nabla (r)\\\\ &=-\vec a\left( \frac{\sin(\vec a\cdot \vec r)}{r^2}\right)-\vec r\left(\frac{2\cos(\vec a\cdot \vec r)}{r^4}\right) \end{align}$$