Consider arbitrary elements $n \in \mathbb N$ and $ \lambda \in \mathbb R$ such that $0 < \lambda < n$.
Problem. My goal is to prove that the supremum
$$ \sup_{r > 0} f(r) $$
is finite, where $f$ is defined as
$$ f(r) = \begin{cases} r^{n-\lambda}\log_2(r+2) \quad &\text{ if } 0 < r \leqslant 1 \\[.2cm] r^{-\lambda}\log_2(r+2) &\text{ if } r > 1.\end{cases} $$
Attempt. I thought about starting by differentiating $f$. Doing so, we obtain
$$ f'(r) = \begin{cases} r^{n-\lambda-1}\left( (n-\lambda)\log_2(r+2) + \frac{r}{r\ln(2) + \ln(4)} \right) \quad &\text{ if } 0 < r \leqslant 1, \\[.2cm] r^{-\lambda - 1}\left( \frac{r}{r\ln(2) + \ln(4)} - \lambda \log_2(r+2)\right) &\text{ if } r > 1.\end{cases} $$
After this, I attempted to determine the zeros of $f'$ but it turns out this isn't an easy task (at least for me). Therefore, I skipped to try and study the signal of the derivative $f'$. It's clear that for $0 < r \leqslant 1$ the derivative $f'$ is always positive, since every member involved in positive.
As for the $r > 1$, to study the signal of $f'$ we must study the expression
$$ \frac{r}{r\ln(2) + \ln(4)} - \lambda \log_2(r+2). $$
Again, just like studying the zeros of the derivative, this turned out to be not so easy. With this in mind, I am requesting help to proceed further.
Thanks for any help in advance.
The question is not asking you to find the supremum, only prove that it is finite. So it is sufficient to prove that $f(r)$ is bounded.
First, while not completely necessary, it's nice to note that $f(r)$ is always positive. As you've noted, for $0 < r \leq 1$, $f(r)$ has positive derivative, so it's increasing.
Then for $r > 1$, we can use Lázaro's hint from the comments that $\lim_{r \rightarrow \infty} \frac{\log r}{r^\lambda} = 0$, and with a little tweaking that gives us that $\lim_{r \rightarrow \infty} f(r) = 0$. Then think back to the original definition of a limit at infinity - we know that for any $M \in \mathbb{R}^+$, there is a value $R$ such that $r \geq R \implies |f(r)| \leq M$.
So pick an $M$. We know that $\sup_{r > R} f(r) < M$. We also know that $f$ is continuous, which means that it is bounded on $(0, R)$, so set $M' = \sup_{r \in (0, R)} f(r)$. Then $\sup_{r > 0} f(r) \leq \max(M, M') < \infty$.