Answer
a)Proved below
b)Proved below
c)$\nabla({r^n})=nr^{n-1}\hat{r}$
Work Step by Step
Separation vector, $\vec{r}=(x-x')\hat{x}+(y-y')\hat{y}+(z-z')\hat{z}$
$|\vec{r}|=r=\sqrt{[(x-x')^2+(y-y')^2+(z-z')^2]}$
a)$\nabla({r^2})=\nabla(\vec{r}\cdot\vec{r})
$$=\frac{\partial}{\partial{x}}[(x-x')^2+(y-y')^2+(z-z')^2]\hat{x}+\frac{\partial}{\partial{y}}[(x-x')^2+(y-y')^2+(z-z')^2]\hat{y}+\frac{\partial}{\partial{z}}[(x-x')^2+(y-y')^2+(z-z')^2]\hat{z}$
$\hspace{1.5cm}=2(x-x')\hat{x}+2(y-y')\hat{y}+2(z-z')\hat{z}=2\vec{r}$
Hence Proved
b)$\nabla\large\Big({\frac{1}{r}}\Big)
$$=\large\frac{\partial}{\partial{x}}\Big(\frac{1}{\sqrt{[(x-x')^2+(y-y')^2+(z-z')^2]}}\Big)\hat{x}+\large\frac{\partial}{\partial{y}}\Big(\frac{1}{\sqrt{[(x-x')^2+(y-y')^2+(z-z')^2]}}\Big)\hat{y} +\large\frac{\partial}{\partial{z}}\Big(\frac{1}{\sqrt{[(x-x')^2+(y-y')^2+(z-z')^2]}}\Big)\hat{z}
=\large-\frac{1}{2}[(x-x')^2+(y-y')^2+(z-z')^2]^{-\frac{3}{2}}\normalsize(2(x-x')\hat{x}+2(y-y')\hat{y}+2(z-z')\hat{z})$
$\hspace{1.7cm}\large=-\frac{1}{2r^3}2\vec{r}=-\frac{\hat{r}}{r^2}$
Hence Proved
c)$\nabla{r^n}=\frac{\partial}{\partial{x}}r^n\hat{x}+\frac{\partial}{\partial{y}}r^n\hat{y}+\frac{\partial}{\partial{z}}r^n\hat{z}=nr^{n-1}(\frac{\partial}{\partial{x}}r\hat{x}+\frac{\partial}{\partial{y}}r\hat{y}+\frac{\partial}{\partial{z}}r\hat{z})$
$\large\frac{\partial}{\partial{x}}r=\frac{1}{2}\frac{1}{\sqrt{[(x-x')^2+(y-y')^2+(z-z')^2]}}2(x-x')$
Similarly finding $\large\frac{\partial}{\partial{y}}r$ and $\large\frac{\partial}{\partial{z}}r$ ,
$\Rightarrow(\frac{\partial}{\partial{x}}r\hat{x}+\frac{\partial}{\partial{y}}r\hat{y}+\frac{\partial}{\partial{z}}r\hat{z})=\frac{1}{2r}2\vec{r}=\hat{r}$
$\therefore\nabla({r^n})=nr^{n-1}\hat{r}$
