My teacher asked to show the following equality: $$\nabla \cdot \vec{F}(r) = \frac{\vec{r}}{r} \cdot \frac{d\vec{F}}{dr} $$
But for me this equality is not valid. For example: $\vec{F}(r)=\frac{1}{r^2} \widehat{r}$
\begin{align} \nabla \cdot \vec{F}(r) &= \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 \frac{1}{r^2}) \\ &= \frac{1}{r^2} \frac{d(1)}{d r} \\ &= 0 \\ \end{align}
And since $F$ depends only on $r$:
$$\frac{\vec{r}}{r} \cdot \frac{d\vec{F}}{dr}= \frac{\vec{r}}{r} \cdot \frac{d \frac{1}{r^2} \widehat{r}}{dr}$$
Now $\frac{\vec{r}}{r}=\widehat{r}$
\begin{align} \frac{\vec{r}}{r} \cdot \frac{d\vec{F}}{dr} &= \widehat{r} \cdot \frac{d \frac{1}{r^2}}{dr}\widehat{r} \\ &= \frac{d \frac{1}{r^2} }{dr} \\ &= -2\frac{1}{r^3} \end{align}
So $$\nabla \cdot \vec{F}(r) \neq \frac{\vec{r}}{r} \cdot \frac{d\vec{F}}{dr}$$
Assuming that $\vec{F}$ is purely radial, i.e. $\vec{F}(r) = F(r) \hat{r}$, you have (product rule for divergence)
$$\nabla \cdot \vec{F}(r) = \nabla \cdot F(r)\hat{r} = F(r) \big(\nabla \cdot\hat{r}\big) + \big(\nabla F(r)\big) \cdot \hat{r} $$
Since the divergence of $\hat{r}$ is $\dfrac{2}{r}$, and the gradient of $F(r)$ is $\dfrac{\partial F(r)}{\partial r} \hat{r}$, the equation above simplifies to $$\nabla \cdot \vec{F}(r) = \dfrac{2F(r)}{r} + \dfrac{\partial F(r)}{\partial r}$$
We can check that for $\vec{F}(r) = \dfrac{1}{r^2}\hat{r}$, both sides now evaluate to $0$.
If in your notation, $\dfrac {\text{d} \vec{F} } {\text{d} r} = \big(\dfrac{2F(r)}{r} + \dfrac{\partial F(r)}{\partial r}\big) \hat{r}$, then everything checks out.