Given this field $$ \mathbf{F(\mathbf{r})}=F_0a^3\left( \frac{2\cos\theta}{r^3}\mathbf{\hat{r}}+\frac{\sin\theta}{r^3}\mathbf{\hat{\theta}}\right) $$ How can I show that the divergence is zero ($r\neq 0$)? This is what I get: $$ \nabla\cdot\mathbf{F}=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2F_r\right)+\frac{1}{r\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta F_\theta\right) $$ The first term I find is $$ -\frac{-2\cos\theta}{r^4} $$ After substitution of $F_r$. And the second term $$ \frac{\sin 2\theta}{r^4\sin\theta} $$ After substition with $F_\theta$. Now, is this part correct and I need to see that they eliminate each other and so the divergence is zero? Or am I doing something wrong when doing the divergence?
When I do the second term, I use that $\sin\theta F_\theta=\frac{\sin^2\theta}{r^3}=\frac{1}{r^3}\frac{1-\cos{2\theta}}{2}$.
We have
$$F_\theta = \frac{\sin \theta}{r^3}$$
So that
$$\frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} (\sin \theta F_\theta) = \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} (\frac{\sin^2 \theta}{r^3}) = \frac{1}{r^4 \sin \theta} \frac{\partial}{\partial \theta} (\sin^2 \theta) = \frac{1}{r^4 \sin \theta} (2\sin \theta \cos \theta) = \frac{2\cos \theta}{r^4} $$
Which should be exactly what you need. Note that
$$ \sin (2\theta) = 2 \sin \theta \cos \theta$$
So this is what you calculated, but you overcomplicated things.