Let $L(x)=\int_1^x \frac1t dt$ prove that $L(xy)=L(x)+L(y),\quad x,y>0$ and some other similar problems. (Without using logarithms)

c) \begin{align*}L(xy)&=\int_1^{xy}\frac1t\,dt\\&=\int_1^x\frac1t\,dt+\int_x^{xy}\frac1t\,dt\end{align*}and so all you have to prove is that $\int_x^{xy}\frac1t\,dt=\int_1^y\frac1t\,dt$. You can do it doing the substitution $t=xu$ and $dt=x\,du$

d) If $n<0$, then $L(2^n2^{-n})=L(1)=0$ and $L(2^n2^{-n})=L(2^n)+L(2^{-n})=L(2^n)+(-n)L(2)$. And $L(2^n)+(-n)L(2)=0\Longleftrightarrow L(2^n)=nL(2)$.

e) Take $y\in[0,+\infty)$. Pick $n\in\mathbb N$ such that $nL(2)>y\Longleftrightarrow L(2^n)>y$. So, by the intermediate value theorem, there is some $x\in[1,2^n]$ such that $L(x)=y$.

On the other hand, if $y\in(-\infty,0]$, $-y=L(x)$ for some $x$ and therefore $y=-L(x)=L\left(\frac1x\right)$.

let $g(x):=f(xy)$ and differentiate with respect to $x$. You arrive at $$g^\prime(x) = f^\prime(xy) y = \frac{1}{xy}y=\frac{1}{x}$$ So $f$ and $g$ satisfy the same differential equation as a function of $x$. This implies the differ by a constant $c(y)$ which depends only on $y$. So

$$c(y) = \int_1^{xy}\frac{1}{t}dt-\int_1^{x}\frac{1}{t}dt =\int_x^{xy}\frac{1}{t}dt$$ No apply the change of variable theorem to show that $c(y)= L(y)$