You are misunderstanding what needs to be shown in order to show that something is an equivalence relation. None of the arguments you present is a valid argument for the proposition you are trying to establish.
Your equivalence relation is defined on the set of positive integers. So if you write $x\sim y$, you are already assuming that $x$ and $y$ are positive integers, there is no need to prove that they are.
The relation is defined as follows: if $x$ and $y$ are positive integers, then
$$x\sim y\Longleftrightarrow \text{there exists a real number }k\text{ such that }y = 3^kx.$$
That means that in order to show that two given numbers $x$ and $y$ are related, then you need to produce a real number $k$ that witnesses the identity $y=3^kx$. For example, to show that $y=9$ and $x=3$ are related, I just need to say: "take $k=1$. Then $9 = 3^1\cdot 3$; that is, $y = 3^1x$, so $x\sim y$ holds." To show that $18\sim 6$, I say "take $k=-1$; then $6 = 3^{-1}(18)$ holds." Etc.
And if you know that $x\sim y$, then you know that there exists a real number $k$ such that $y=3^kx$.
Thus, to show that $\sim$ is reflexive, you need to show that given any positive integer $x$, you can find a real number $k$ (which may depend on $x$) such that $x = 3^k x$. You need to say who $k$ is. Your argument about "being positive when $k\gt 0$" doesn't get you there in any way.
To show that $\sim$ is symmetric, you have to show that if you already know that $x\sim y$, so that you know there is a real number $k$ such that $y=3^k x$, then you can find some real number $\ell$ such that $x=3^{\ell}y$. This will witness the fact that $y\sim x$ holds, showing symmetry. You already know that $x$ and $y$ are positive integers. You know that simply because you know that $x\sim y$ holds, and that means that $x$ and $y$ have to be positive integers in the first place.
To show that $\sim$ is transitive, you have to show that if you already know $x\sim y$ and $y\sim z$, then you can exhibit a real number $r$ such that $z=3^rx$; this will witness $x\sim z$. You know there is a real number $k$ such that $y=3^kx$ (because $x\sim y$); and you know there is a real number $\ell$ such that $z=3^{\ell}y$ (because $y\sim z$); now you need to produce that number $r$ somehow. Again, you already know that $x$, $y$, and $z$ are positive integers, because you already know that $x\sim y$ and $y\sim z$ are true, which means, inter alia, that $x$, $y$, and $z$ are all positive integers.
I’ll get you started. To show that $\sim$ is reflexive, you must show that if $n\in\Bbb N$, then $n\sim n$. Check the definition of $\sim$: this means that $n\cdot n$ is a square. Of course $n\cdot n=n^2$ is a square, so $n\sim n$, and $\sim$ is reflexive. You should have no trouble showing that $\sim$ is symmetric. For transitivity, suppose that $k\sim m$ and $m\sim n$. Then $km$ and $mn$ are squares, say $km=a^2$ and $mn=b^2$; you must show that $k\sim n$, i.e., that $kn$ is a square. Try to write $kn$ in terms of the pieces that you already have, doing it in a way that demonstrates that $kn$ is a square.
$[3]$ is by definition the set of all $n\in\Bbb N$ such that $3\sim n$, i.e., such that $3n$ is a square. What does this tell you about $n$? HINT: What can you say about the number of factors of $3$ in the prime factorization of $n$? Thinking in similar terms will get you through the rest of (b) as well.
For (c) you should be thinking about the prime factorizations of $a$ and $b$.
Best Answer
You’re right about a and b but not c. Take $x=\sqrt{2}$,$y =-\sqrt{2}$, $z =\sqrt{2}$. Then $x \sim y$, $y\sim z$ but $x+z = 2\sqrt{2}$ is irrational.