Given: $$x\, f(x) = \log(x) \qquad \forall\; x > 0$$ we need to prove that the $n^{th}$ derivative of $\,f(x)\,$ at $\,x = 1\,$ is: $$f^{(n)}(1) = (-1)^{n+1}\, n! \, \left( 1 + \frac{1}{2} + \ldots + \frac{1}{n} \right)$$
Proved for $\,n=1, 2\,$ and tried applying mathematical induction, wherefrom I am getting $\,f^{(n)}(x)\,$ from $\,f^{(n-1)}(x)$,$\:$ but $\,f^{(n)}(1)\,$ seems underivable from $\,f^{(n-1)}(1)$.
The suggestive form in which the function is written can make things a bit neater, rather than just taking a bunch of derivatives of $\ln(x)/x$ (which could be messy).
So I'll take one derivative of both sides of your equation, using the product rule on the left side, and get $$xf'(x)+f(x)=\frac{1}{x}.$$ That tells you what $f'$ is. Take another derivative to get $$xf''(x)+f'(x)+f'(x) = xf''(x)+2f'(x) = -\frac{1}{x^2},$$ which allows you to solve for $f''$ in terms of $f'$. Let's try once more to try to see a pattern. We find $$xf^{(3)}(x)+f''(x)+2f''(x)=xf^{(3)}(x)+3f^{(2)}(x)=\frac{2}{x^3}.$$ Maybe one more derivative gives $$xf^{(4)}(x)+4f^{(3)}(x) = -\frac{2 \cdot 3}{x^4}.$$ At this point, I'd be comfortable positing that the following is true for all $n \geq 1$: $$xf^{(n)}(x)+nf^{(n-1)}(x)=(-1)^{n-1}\frac{(n-1)!}{x^n}$$ You should be able to prove this easily by induction. Plugging in $x=1$, you can then use this to prove your claim by induction.
Another proof, by induction. Let $f(x)=\log(x)/x$, then:
$$f^{(n)}(x)=(-1)^{n+1}n!\frac{H_n-\log x}{x^{n+1}}$$
Where $H_n=\sum_{k=1}^n \frac1k$.
The formula is trivially true for $n=0$ (the sum in $H_0$ is empty, hence has value $0$). Suppose it's true for $n\ge0$, then:
$$f^{(n+1)}(x)=-(n+1)(-1)^{n+1}n!\frac{H_n}{x^{n+2}}-(-1)^{n+1}n!\frac{1}{x^{n+2}}+(n+1)(-1)^{n+1}n!\frac{\log x}{x^{n+2}}\\=(-1)^{n+2}(n+1)!\frac{H_n+\frac{1}{n+1}-\log x}{x^{n+2}}=(-1)^{n+2}(n+1)!\frac{H_{n+1}-\log x}{x^{n+2}}$$
Hence the formula is true for $n+1$, hence, by induction, for all $n\ge0$.
Then
$$f^{(n)}(1)=(-1)^{n+1}n!H_{n}$$
Since you only need the value at $x=1$, no need to bother with the algebra of multiple differentiations and the Leibniz rule. You can instead use a Taylor series around 1: $$\begin{split} \frac{\log (1+h)} {1+h} &=\left ( \sum_{n\geq 1} (-1)^{n+1}\frac {h^n}n\right)\left ( \sum_{n\geq 0} (-1)^{n}h^n\right)\\ &= \sum_{n\geq 1}\left((-1)^{n+1}\sum_{k=1}^n\frac{1}{k}\right)h^n\\ &= \sum_{n\geq 1}\frac{f^{(n)}(1)}{n!}h^n \end{split}$$ Identifying the coefficients gives the result.
$(uv)^{(n)}=\sum\limits_{k=0}^{n} \binom {n} {k} u^{(k)}v^{(n-k)}$ and it is easy to write down all the derivatives of $u =\log x$ and $v=\frac 1 x$.
For this formula (Leibniz Rule) See inhttps://www.math24.net/leibniz-formula/
Instead we could find the $n$th derivative of
$$g(x) = f(x+1) = \frac{\log(1+x)}{1+x}$$
at $x=0$. We have that
$$xg(x) + g(x) = g(0)+\sum_{n=1}^{\infty} \left[\frac{g^{(n)}(0)+ng^{(n-1)}(0)}{n!}\right]x^n = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}x^n$$
which gives us the recurrence relation
$$\begin{cases}g^{(n)}(0)+ng^{(n-1)}(0) = (-1)^{n+1}(n-1)! \\ g^{(1)}(0) = 1 \\ \end{cases}$$
The recurrence relation can be solved by making your ansatz exactly the proposed solution.
