given function $f(n)$ such then:
$$\lim_{n \rightarrow \infty}{f(n)}=\infty$$
It's correct that $\forall k \in \mathbb{N}$ exists:
$$\lim_{n \rightarrow \infty}{\frac{f(n^k)}{f(n^{k+1})}}=0$$
?
I don't have idea how to prove that. But in other hand, I can't find example that prove that it's not correct.
In addition, if it's not must be equal to $0$.
It's correct that $\forall k \in \mathbb{N}$ exists:
$$\lim_{n \rightarrow \infty}{\frac{f(n^k)}{f(n^{k+1})}}<1$$
?
No. Take for example $f(x)=\ln(x)$ then for $k\in\mathbb{N}$, $$\lim_{n \rightarrow \infty}{\frac{f(n^k)}{f(n^{k+1})}}= \lim_{n \rightarrow \infty}{\frac{\ln(n^k)}{\ln(n^{k+1})}}= \lim_{n \rightarrow \infty}{\frac{k\ln(n)}{(k+1)\ln(n)}}=\frac{k}{k+1}.$$
As regards your second question, note that the limit (if exists) is $\leq 1$ if $f$ is increasing.
However the limit can be $>1$. For $k\geq 2$, let us define $f(n)=2\ln(n)$ if $n$ is a perfect $k$-power and $f(n)=\ln(n)$ otherwise then $$\limsup_{n \rightarrow \infty}{\frac{f(n^k)}{f(n^{k+1})}}=\frac{2k}{k+1}>1.$$
No, in general you cannot conclude. It is true, for instance, for $f(x)=x$, but $f(x) = \log(x)$ yields
$$ \lim_{n \to \infty} \frac{\log(n^{k})}{\log(n^{k+1})} = \lim_{n \to \infty} \frac{k\log(n)}{(k+1)\log(n)} = \frac{k}{k+1}.$$
