$\vec{R}=x \hat{i} + y\hat{j} + z\hat{k}$ and $r=|\vec{R}|=\sqrt{x^2+y^2+z^2}$
Prove that $\nabla^2f(r)=f''(r)+\frac{2}{r}f'(r)$
So we need to basically show that $$\frac{\partial^2}{\partial x^2}f(r)+\frac{\partial^2}{\partial y^2}f(r)+\frac{\partial^2}{\partial z^2}f(r)=\frac{d^2f}{dr^2}+\frac{2}{r}\frac{df}{dr}$$
$f(r)$ is a scalar function of $r$.
I'm not sure how to go about proving this. How can we possibly express $\frac{d^2f}{dr^2}$ and $\frac{df}{dr}$ in terms of $\frac{\partial^2}{\partial x^2}f(r)$, $\frac{\partial^2}{\partial y^2}f(r)$ and $\frac{\partial^2}{\partial z^2}f(r)$ ?
By using the chain rule, we have $$ \begin{aligned} \frac{\partial f}{\partial x}=\frac{d f}{dr}\cdot\frac{\partial r}{\partial x} \end{aligned} $$ and $$ \begin{aligned} \frac{\partial^2 f}{\partial x^2}&=\frac{\partial}{\partial x}\left(\frac{d f}{dr}\right)\cdot\frac{\partial r}{\partial x}+\frac{d f}{dr}\cdot\frac{\partial ^2r}{\partial x^2}\\ &=\frac{\partial^2f}{\partial r^2}\cdot\left(\frac{\partial r}{\partial x}\right)^2+\frac{d f}{dr}\cdot\frac{\partial ^2r}{\partial x^2}\\ &=\frac{\partial^2f}{\partial r^2}\cdot\left(\frac{x}{r}\right)^2+\frac{d f}{dr}\cdot\left(\frac{1}{r}-\frac{x^2}{r^3}\right) \end{aligned} $$ Thus, $$ \begin{aligned} \frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}+\frac{\partial^2 f}{\partial z^2}&=\frac{\partial^2f}{\partial r^2}\cdot\left[\left(\frac{x}{r}\right)^2+\left(\frac{y}{r}\right)^2+\left(\frac{z}{r}\right)^2\right]\\ &+\frac{d f}{dr}\cdot\left(\frac{3}{r}-\frac{x^2+y^2+z^2}{r^3}\right)\\ &=\frac{\partial^2f}{\partial r^2}+\frac{2}{r}\frac{df}{dr} \end{aligned} $$