Consider following integral:
$$F(x) = \int \frac{1}{x}\cdot \frac{1}{ln(x)}dx$$
which we can substitute with $u = u(x) = ln(x)$:
$$\int \frac{1}{x} \frac{1}{u}dx$$
then we can find $dx$:
$$u'(x) = \frac{du}{dx} = \frac{1}{x} \iff dx = \frac{x}{du}$$
so we can insert this into our substituted integral:
$$\int \frac{1}{x}\frac{1}{u} \frac{x}{du}$$
this allows us to get rid of $\frac{1}{x}$:
$$\int \frac{1}{u} \frac{1}{du}$$
but it leaves us with $\frac{1}{du}$ which is no valid integral expression.
I also tried out to set $u = x \cdot ln(x)$ which yields in $\frac{ln(x \cdot ln(x))}{ln(x)+1} + C$. Consulting wolframalpha shows that this is not a valid answer also they substituted $u = ln(x)$ as I tried.
Your this step is wrong: $$dx = \frac{x}{du}\tag{wrong}$$ Instead it should be: $$dx=xdu\tag{right}$$ $$f(x)=\int\frac{dx}{x\ln x}=\int\frac{d(\ln x)}{\ln x}=\ln(\ln x)+c$$