# Solución

Como $r = \sqrt{x^2 + y^2 + z^2}$, la regla del cociente nos da

$$\begin{aligned} \frac{\partial}{\partial x} \bigg(\frac{x}{r^3}\bigg) &= \frac{\partial}{\partial x} \bigg(\frac{x}{\big(x^2 + y^2 + z^2\big)^{3/2}}\bigg)\\ &= \frac{\big(x^2 + y^2 + z^2\big)^{3/2}- x\bigg[ \frac32\big(x^2 + y^2 + z^2\big)^{1/2} 2x\bigg]}{\big(x^2 + y^2 + z^2\big)^{3}}\\ &= \frac{r^3-3x^2r}{r^6} = \frac{r^2-3x^2}{r^5} \end{aligned}$$

Similarmente

$$\frac{\partial}{\partial y} \bigg(\frac{y}{r^3}\bigg) = \frac{r^2-3y^2}{r^5}\;\;\text{y }\;\;\frac{\partial}{\partial z} \bigg(\frac{z}{r^3}\bigg) = \frac{r^2-3z^2}{r^5}$$

Por lo tanto

$$\begin{aligned} div\;\bold{F}_r &= \frac{r^2-3x^2}{r^5} + \frac{r^2-3y^2}{r^5} + \frac{r^2-3z^2}{r^5}\\ &= \frac{3r^2 - 3\big(x^2+y^2+z^2\big)}{r^5}\\ &= \frac{3r^2-3r^2}{r^5} = 0 \end{aligned}$$