$n^{th}$ derivative of $\log(x)/x$ The problem
Given: $$x\, f(x) = \log(x) \qquad \forall\; x &gt; 0$$
we need to prove that the $n^{th}$ derivative of $\,f(x)\,$ at $\,x = 1\,$ is: $$f^{(n)}(1) = (-1)^{n+1}\, n! \, \left( 1 + \frac{1}{2} + \ldots + \frac{1}{n} \right)$$
What I tried
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.
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.