Answer
$$\nabla f =F^{\prime}(r) e_{\mathbf{r}}$$
Work Step by Step
Since
\begin{aligned}
\frac{\partial f}{\partial x}&=F^{\prime}(r) \frac{\partial r}{\partial x}\\
&=F^{\prime}(r) \cdot \frac{2 x}{2 \sqrt{x^{2}+y^{2}+z^{2}}}\\
&=F^{\prime}(r) \cdot \frac{x}{r}\\
\frac{\partial f}{\partial y}&=F^{\prime}(r) \frac{\partial r}{\partial y}\\
&=F^{\prime}(r) \cdot \frac{2 y}{2 \sqrt{x^{2}+y^{2}+z^{2}}}\\
&=F^{\prime}(r) \cdot \frac{y}{r}\\
\frac{\partial f}{\partial z}&=F^{\prime}(r) \frac{\partial r}{\partial z}\\
&=F^{\prime}(r) \cdot \frac{2 z}{2 \sqrt{x^{2}+y^{2}+z^{2}}}\\
&=F^{\prime}(r) \cdot \frac{z}{r}
\end{aligned}
Then
\begin{align*}
\nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle\\
&=\left\langle F^{\prime}(r) \frac{x}{r}, F^{\prime}(r) \frac{y}{r}, F^{\prime}(r) \frac{z}{r}\right\rangle\\
&=\frac{F^{\prime}(r)}{r}\langle x, y, z\rangle\\
&= F^{\prime}(r) \frac{\mathbf{r}}{\|\mathbf{r}\|}\\
&=F^{\prime}(r) e_{\mathbf{r}}
\end{align*}
