n-th derivative of $\frac{\ln x}{x}$.

Let $g(x)=f(x+1)$ so that $g^{(n)}(0)=f^{(n)}(1)$.

Then since $\log(1+x) = \int dx/(1+x) =\sum_{n=0}^\infty {(-1)^n x^{n+1}/ (n+1)}$ $$g(x)={\log(1+x)\over1+x}=\sum_{n=0}^\infty(-1)^nx^n \sum_{n=0}^\infty {(-1)^n x^{n+1} \over n+1}=\sum_{n=0}^\infty(-1)^nx^n \sum_{m=1}^\infty {(-1)^{m-1} x^{m} \over m} = \sum_{n=1}^\infty\left(\sum_{k=1}^n{(-1)^{n-1} \over k}\right) x^n$$ and apply Taylor's formula (as well as the identity $(-1)^{n-1}=(-1)^{n+1}$).

Alternate I

Let $I(x) = \int f(1+x) \; dx = {1\over2}(\log(1+x))^2$, so that $I^{(n+1)}(0)=f^{(n)}(1)$. Then $$I(x) = {1\over2}\left(\sum_{m=1}^\infty {(-1)^{m-1} x^{m} \over m}\right)^2 =\sum_{m=2}^\infty\left(\sum_{k=1}^m{(-1)^{m-2}\over2k(m-k)}\right)x^n =\sum_{m=2}^\infty\left(\sum_{k=1}^m{(-1)^{m-2}\over2m}\left({1\over k}+{1\over m-k}\right)\right)x^m\,,$$ which, since $k$ and $m-k$ each run over the integers $1,2,\dots,m$ in the inside sum, is equal to $$=\sum_{m=2}^\infty\left(\sum_{k=1}^m{(-1)^{m-2}\over mk}\right)x^m =\sum_{n=1}^\infty\left(\sum_{k=1}^{n+1}{(-1)^{n-1}\over(n+1)k}\right)x^{n+1}$$ and again apply Taylor's formula.

Alternate II

It looks so much like an impossible induction problem, I could not resist trying...

Let $y=f(x)=(\log(x)/x)$, so that the derivative $D(xy) = x\,y'+y = D(\log x) = 1/x$. By induction then, we have that the $n^{th}$ derivative is given by $D^n(xy) = x\,y^{(n)}+n\,y^{(n-1)} = (-1)^{n-1} (n-1)!/x^n$, so that at $x=1$, we have the equation $$y^{(n)}+n\,y^{(n-1)} = (-1)^{n-1} (n-1)!\,.$$

Now we're ready to prove the formula for $f^{(n)}(1)$. That $f'(1)=1$ is easily verified. Assume that at $x=1$, the $n^{th}$ derivative $y^{(n)}=f^{(n)}(1)$ satisfies $$y^{(n)}=(-1)^{n-1}(n!)(1+\frac{1}{2}+\cdots+\frac{1}{n})\,.$$ Then applying the equation above for the $n+1^{st}$ derivative, we have $$y^{(n+1)}=(-1)^{n} n! -(n+1)y^{(n)} ={(-1)^{n} (n+1)!\over n+1}-(n+1){(-1)^{n-1}(n!)(1+\frac{1}{2}+\cdots+\frac{1}{n})} ={(-1)^{n}(n+1)!(1+\frac{1}{2}+\cdots+\frac{1}{n}+\frac{1}{n+1})}$$ So by induction the formula for the derivative at $x=1$ is established.

Using the product rule with $g(x)=1/x$ and $f(x)=\ln(x)$, we have

$$ f^{(n)}(x)=\sum_{i=0}^{n} {n\choose i} g^{(i)}(x)h^{(n-i)}(x)$$

$$= (-1)^n n!\, x^{-(n+1)}\ln(x) +\sum_{i=0}^{n-1} {n\choose i}(-1)^i i!\, x^{-i-1}(-1)^{n-i-1}(n-i-1)!\,x^{-n+i} $$

$$=(-1)^n n!\, x^{-(n+1)}\ln(x) + (-1)^{n-1}x^{-n-1}\sum_{i=0}^{n-1} {n\choose i}i!(n-i-1)! $$

$$ =(-1)^n n!\, x^{-(n+1)}\ln(x) + (-1)^{n-1}x^{-n-1}\sum_{i=0}^{n-1} \frac{n!}{i!(n-i)!}i!(n-i-1)! $$

$$\implies f^{(n)}(x) = (-1)^n n!\, x^{-(n+1)}\ln(x) +(-1)^{n-1}n!\,x^{-n-1}\sum_{i=0}^{n-1} \frac{1}{(n-i)} .$$

Now, you can substitute $x=1$ to get the desired answer.