For $$H_{k}(x)=\frac{(-1)^{k}}{\sqrt{k!}}\exp\left\{\frac{x^{2}}{2}\right\}\frac{d^{k}}{dx^{k}}\exp\left\{-\frac{x^{2}}{2}\right\}$$
I want to prove $H'_{k}(x)=\sqrt{k}H_{k-1}(x)$.
So far I have $$\begin{aligned} H'_{k}(x)&=x\frac{(-1)^{k}}{\sqrt{k!}}\exp\left\{\frac{x^{2}}{2}\right\}\frac{d^{k}}{dx^{k}}\exp\left\{-\frac{x^{2}}{2}\right\}+\frac{(-1)^{k}}{\sqrt{k!}}\exp\left\{\frac{x^{2}}{2}\right\}\frac{d^{k+1}}{dx^{k+1}}\exp\left\{-\frac{x^{2}}{2}\right\} \\ &=x\frac{(-1)^{k-1}}{\sqrt{k!}}\exp\left\{\frac{x^{2}}{2}\right\}\frac{d^{k-1}}{dx^{k-1}}\left(x\exp\left\{-\frac{x^{2}}{2}\right\}\right)+\frac{(-1)^{k-1}}{\sqrt{k!}}\exp\left\{\frac{x^{2}}{2}\right\}\frac{d^{k}}{dx^{k}}\left(x\exp\left\{-\frac{x^{2}}{2}\right\}\right)\\ &=x\sqrt{k}\frac{(-1)^{k-1}}{\sqrt{(k-1)!}}\exp\left\{\frac{x^{2}}{2}\right\}\frac{d^{k-1}}{dx^{k-1}}\left(x\exp\left\{-\frac{x^{2}}{2}\right\}\right)+\sqrt{k}\frac{(-1)^{k-1}}{\sqrt{(k-1)!}}\exp\left\{\frac{x^{2}}{2}\right\}\frac{d^{k}}{dx^{k}}\left(x\exp\left\{-\frac{x^{2}}{2}\right\}\right) \end{aligned}$$
I'm having trouble dealing with the initial $x$ and also with the derivatives.
The typical way to get at this is through the generating function. \begin{align} \sum_{n=0}^{\infty}\frac{(-t)^n}{n!}\frac{d^{n}}{dx^{n}}e^{-x^2/2} & = e^{-(x-t)^2/2} \\ \sum_{n=0}^{\infty}\frac{(-t)^{n}}{n!}e^{x^{2}/2}\frac{d^{n}}{dx^{n}}e^{-x^2/2} & =e^{x^{2}/2-(x-t)^2/2}=e^{xt-t^2} \\ \sum_{n=0}^{\infty}\frac{t^{n}}{\sqrt{n!}}H_{n}(x)& =e^{xt-t^2} \end{align} Therefore, \begin{align} \sum_{n=0}^{\infty}\frac{t^{n}}{\sqrt{n!}}H_{n}'(x) & = \frac{d}{dx}e^{xt-t^2} = te^{xt-t^{2}} \\ & = \sum_{n=0}^{\infty}\frac{t^{n+1}}{\sqrt{n!}}H_{n}(x) \\ & = \sum_{n=1}^{\infty}\frac{t^{n}}{\sqrt{(n-1)!}}H_{n-1}(x) \\ & = \sum_{n=1}^{\infty}\frac{t^{n}}{\sqrt{n!}}\sqrt{n}H_{n-1}(x). \end{align} So $H_{n}'(x)=\sqrt{n}H_{n-1}(x)$ for $n \ge 1$ and $H_{0}'(x)=0$.
For the other identity you mentioned, you can view $H_n$ as a differential operator acting on the constant function $1$: \begin{align} H_n(x)&=\frac{(-1)^n}{\sqrt{n!}}\left[e^{x^2/2}\frac{d^n}{dx^n}e^{-x^2/2}\right]1 \\ &=\frac{(-1)^n}{\sqrt{n!}}\left[e^{x^2/2}\frac{d}{dx}e^{-x^2/2}\right]^n1 \\ &=\frac{(-1)^n}{\sqrt{n!}}\left[\frac{d}{dx}-x\right]^n1 \end{align} Therefore, \begin{align} H_{n}'(x)&=\left[\frac{d}{dx}-x\right]H_n(x)+xH_{n}(x) \\ &=\frac{(-1)^{n}}{\sqrt{n!}}\left[\frac{d}{dx}-x\right]^{n+1}1+xH_{n}(x) \\ &=-\sqrt{n+1}\frac{(-1)^{n+1}}{\sqrt{(n+1)!}}\left[\frac{d}{dx}-x\right]^{n+1}1+xH_{n}(x) \\ &=-\sqrt{n+1}H_{n+1}(x)+xH_{n}(x) \end{align} The generating function method will also work, but, in this case, nothing could be more transparent than the operator method, at least from my point of view. Now if you differentiate and use the first identity, \begin{align} H_{n}''(x) & = -\sqrt{n+1}H_{n+1}'(x)+H_{n}(x)+xH_{n}'(x) \\ &= -\sqrt{n+1}\sqrt{n+1}H_{n}+H_{n}(x)+xH_{n}'(x) \\ &= -nH_{n}(x)+xH_{n}'(x) \end{align} Hence, $$ H_{n}''(x)-xH_{n}'+nH_{n}=0. $$