The only problem is that you’re looking at the wrong three points: you’re looking at $x+2h,x+h$, and $x$, and the version that you want to prove is using $x+h,x$, and $x-h$. Start with $$f\,''(x)=\lim_{h\to 0}\frac{f\,'(x)-f\,'(x-h)}h\;,$$ and you’ll be fine.
To see that this really is equivalent to looking at $$f\,''(x)=\lim_{h\to 0}\frac{f\,'(x+h)-f\,'(x)}h\;,$$ let $k=-h$; then
$$\begin{align*}
f\,''(x)&=\lim_{h\to 0}\frac{f\,'(x)-f\,'(x-h)}h\\
&=\lim_{-k\to0}\frac{f\,'(x)-f\,'(x-(-k))}{-k}\\
&=\lim_{k\to 0}\frac{f\,'(x-(-k))-f\,'(x)}k\\
&=\lim_{k\to 0}\frac{f\,'(x+k)-f\,'(x)}k\;,
\end{align*}$$
and renaming the dummy variable back to $h$ completes the demonstration.
Hint: We can find an elaboration of the $n$-th derivative of $x^x$ in an example (p. 139) of Advanced Combinatorics by L. Comtet. The idea is based upon a clever Taylor series expansion. Using the differential operator $D_x^j:=\frac{d^j}{dx^j}$ the following holds:
The $n$-th derivative of $x^x$ is
\begin{align*}
D_x^n x^x=x^x\sum_{i=0}^n\binom{n}{i}(\ln(x))^i\sum_{j=0}^{n-i}b_{n-i,n-i-j}x^{-j}\tag{1}
\end{align*}
with $b_{n,j}$ the Lehmer-Comtet numbers.
These numbers follow the recurrence relation
\begin{align*}
b_{n+1,j}=(j-n)b_{n,j}+b_{n,j-1}+nb_{n-1,j-1}\qquad\qquad n,j\geq 1
\end{align*}
and the first values, together with initial values are listed below.
\begin{array}{c|cccccc}
n\setminus k&1&2&3&4&5&6\\
\hline
1&1\\
2&1&1\\
3&-1&3&1\\
4&2&-1&6&1\\
5&-6&0&5&10&1\\
6&24&4&-15&25&15&1\\
\end{array}
The values can be found in OEIS as A008296. They are called Lehmer-Comtet numbers and were stored in the archive by N.J.A.Sloane by referring precisely to the example we can see here.
Example: $n=2$
Let's look at a small example. Letting $n=2$ we obtain from (1) and the table with $b_{n,j}$:
\begin{align*}
D_x^2x^x&=x^x\sum_{i=0}^2\binom{2}{i}(\ln(x))^i\sum_{j=0}^{2-i}b_{2-i,2-i-j}x^{-j}\\
&=x^x\left(\binom{2}{0}\sum_{j=0}^2b_{2,2-j}x^{-j}+\binom{2}{1}\ln(x)\sum_{j=0}^1b_{1,1-j}x^{-j}\right.\\
&\qquad\qquad\left.+\binom{2}{2}\left(\ln(x)\right)^2\sum_{j=0}^0b_{0,0-j}x^{-j}\right)\\
&=x^x\left(\left(b_{2,2}+b_{2,1}\frac{1}{x}+b_{2,0}\frac{1}{x^2}\right)+2\ln(x)\left(b_{1,1}+b_{1,0}\frac{1}{x}\right)
+(\ln(x))^2b_{0,0}\right)\\
&=x^x\left(1+\frac{1}{x}+2\ln(x)+\left(\ln(x)\right)^2\right)
\end{align*}
in accordance with the result of Wolfram Alpha.
Note: A detailed answer is provided in this MSE post.
Another variation of the $n$-th derivative of $x^x$ is stated as Remark 7.4 in section 7.8 of Combinatorial Identities, Vol. 5 by H.W. Gould.
Let $\{F_x(k)\}_{k=0}^\infty$ be a sequence with $F_0(x)=1, F_1(x)=1$, and
\begin{align*}
F_k(x)=-D_xF_{k-1}(x)+\frac{k-1}{x}F_{k-2}(x)\qquad k\geq 2.
\end{align*}
Assuming $x\neq 0$ the following holds
\begin{align*}
D_x^n(x^x)=x^x\sum_{k=0}^n(-1)^k\binom{n}{k}(1+\ln x)^{n-k}F_k(x)
\end{align*}
Best Answer
There are formulas for $n$-th derivate for specific functions where you can find a pattern. There is no special source to link since it is something super general. Everyone does it with several functions.
I will let you here some of them.
EXAMPLE $1$:
$$f(x)=x\cdot e^{-x}$$ $$f^{n)}(x)=((-1)^n)(x\cdot e^{-x})+((-1)^{n+1}\cdot n)(e^{-x})$$
EXAMPLE $2$:
$$g(x)=\frac{1}{1-x}$$ $$g^{n)}(x)=\frac{n!}{(1-x)^{n+1}}$$
Yours is one more example. It is correct. You have found a pattern and you have defined the $n$-th derivate for your function.