Is always correct statement that if natural numbers $a,b \in \Bbb N$ for which LCM$(a,b)=16\cdot(a,b)$, then $a|b$ or $b|a$?
I used formula that LCM$(a,b)=\frac{a\cdot b}{(a,b)}$
$\frac{a\cdot b}{(a,b)}=16\cdot(a,b) \implies a\cdot b= (4\cdot(a,b))^2$
Is it somehow useful? What should I do next?
You're nearly there!
Let $ a = k_a (a,b) , b = k_b (a,b)$ where $(k_a, k_b ) = 1$.
Hint: What is the value of $ k_a k_b$ according to your equation?
Hint: Can we conclude that one of them must be 1?
If yes, then we either have $ a \mid b$ or $b\mid a$. If no, then we have a counterexample.
Let $d= (a,b)$. We know that $LCM (a,b) ={ab\over d}$. Then as you noted we have $$ab =16d^2$$ We can also write $a=dx$ and $b=dy$ where $x,y$ are coprime. So $$xy=16$$ Now it is not difficult to finish.
Hint $ $ Conceptually it is obvious: a prime $\,p\,$ occurs in $\,c = {\rm lcm}(a,b)/\gcd(a,b)\!\iff\! p\, $ occurs to different powers in $\,a\,$ and $\,b.\,$ Since only the prime $\,p=2\,$ occurs in OP's $\,c = 2^4,\,$ we deduce that $\,a,b\,$ have prime factorizations that differ only in the power of the prime $\,2,\,$ so the one with the least power of $\,2\,$ divides the other (we used existence and uniqueness of prime factorizations throughout). Hence it is clear that the result still holds true if we generalize $\,16=2^4\to \color{#c00}{p^k}\,$ any prime power.
Or cancelling their gcd reduces to case $\,a,b\,$ coprime so $\,{\rm lcm}(a,b)\!=\!ab = \color{#c00}{p^k}\smash{\overset{\rm wlog}\Rightarrow}^{\phantom{|^l}}\! a=1\,$ so $\,a\mid b$ (note the divisibilities are not affected by gcd cancellation since $\,ad\mid bd\iff a\mid b).\, $
Homogeneous reduction For statements similarly homogeneous in $\,a,b\,$ we can cancel (a power of) the gcd to reduce to the case $\,a,b\,$ coprime, e.g. see here and here and here for further examples. In more advanced contexts one doesn't explicitly change variables and cancel; rather one simply writes: "being homogeneous in $\,a,b\,$ wlog we may reduce to the case $\,a,b\,$ coprime $\ldots$".
