This is from my mathematical physics book. I don't know how the right side arrived from the left side of the equation: $$x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} + z\frac{\partial f}{\partial z} = \frac{df}{dr}\vec r \cdot\nabla r$$
Note that $f$ is a function of $r = \sqrt{x^2 + y^2 + z^2}$ alone: $f(r)$.
Notice
$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial r} \frac{\partial r}{\partial x}$$
by the chain rule. Since $f$ is just a function of $r$, $\partial f/\partial r = df/dr$. Using analogous results for y and z,
$$x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} + z \frac{\partial f}{\partial z} = \frac{df}{dr} (\frac{\partial r}{\partial x} x + \frac{\partial r}{\partial y} y + \frac{\partial r}{\partial z} z)$$
Since $\vec{r} = (x,y,z)$ and $\nabla r = (\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial r}{\partial z})$, this can be written more compactly as
$$ x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} + z \frac{\partial f}{\partial z} = \frac{df}{dr} (\frac{\partial r}{\partial x} x + \frac{\partial r}{\partial y} y + \frac{\partial r}{\partial z} z) = \frac{df}{dr} \vec{r} \cdot \nabla r$$