I would do this using one of the two following methods:
First Method:
Write your vector field in Cartesian coordinates ($x,y,z$). The $\vec{r}$-vector is:
$$\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}=
\begin{pmatrix}
x\\y\\z
\end{pmatrix},
$$
where the hat-vectors are the unit vectors (you might know them as $(\hat{i},\hat{j},\hat{k})$ instead). Next is the $r^3$:
$$r^3=(r^2)^{3/2}=\left(x^2+y^2+z^2 \right)^{3/2}$$
If we define the vector operator $\vec{\nabla}$ as:
$$\vec{\nabla}=
\begin{pmatrix}
\frac{\partial}{\partial x}\\ \frac{\partial}{\partial y}\\ \frac{\partial}{\partial z}
\end{pmatrix},
$$
The divergence of a general vector field $\vec{F}=
\begin{pmatrix}
F_x\\F_y\\F_z
\end{pmatrix}
$ can be written neatly as:
$$\text{div}(\vec{F})=\vec{\nabla}\bullet\vec{F}=\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z}$$
Remember, this formula is only true in Cartesian coordinates and, in particular, not in spherical coordinates.
Now to find your specific divergence:
$$
\begin{align}
\text{div}\left(\vec{r}/r^3 \right)&=\vec{\nabla}\bullet \left( \frac{1}{\left(x^2+y^2+z^2 \right)^{3/2}}
\begin{pmatrix}
x\\y\\z
\end{pmatrix}\right)\\&=\frac{\partial}{\partial x}\left(\frac{x}{\left(x^2+y^2+z^2 \right)^{3/2}}\right)+\frac{\partial}{\partial y}\left(\frac{y}{\left(x^2+y^2+z^2 \right)^{3/2}}\right)+\frac{\partial}{\partial z}\left(\frac{z}{\left(x^2+y^2+z^2 \right)^{3/2}}\right)
\\
&=3\left(x^2+y^2+z^2 \right)^{-3/2}-3(x^2+y^2+z^2)\left(x^2+y^2+z^2\right)^{-5/2}=0
\end{align}
$$
The fact that $r\neq 0$ was used to cancel the parentheses here and there (and of course to allow you to divide by $r^3$ to begin with).
Second Method:
We will now work in spherical coordinates, where your vector field is:
$$\vec{F}=
\begin{pmatrix}
F_r\\F_{\theta}\\F_{\phi}
\end{pmatrix}
=
\begin{pmatrix}
1/r^2\\0\\0
\end{pmatrix}
$$
Here, $\theta$ is the polar angle and $\phi$ is the azimuthal angle. It is already looking much simpler. We only divide by $r^2$ since that is the actual field in the $\hat{r}$-direction. The extra $r$-factor in $r^3$ comes from normalizing the vector $\vec{r}=r\hat{r}$, thus giving $\vec{r}/r^3=\hat{r}r/r^3=\hat{r}/r^2$.
The general divergence in spherical coordinates is:
$$\text{div}(\vec{F})=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2F_r)+\frac{1}{r\sin(\theta)}\frac{\partial}{\partial \theta}(F_{\theta}\sin(\theta))+\frac{1}{r\sin(\theta)}\frac{\partial F_{\phi}}{\partial \phi}$$
Okay, now it is not looking that simple anymore. But notice that only the first term is non-zero, since $F_{\theta}=F_{\phi}=0$. Thus:
$$\text{div}(\vec{F})=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2F_r)=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2r^{-2})$$
Since $r\neq 0$:
$$\text{div}(\vec{F})=\frac{1}{r^2}\frac{\partial}{\partial r}(1)=0$$
