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For R to be reflexive, we need $(x, x) \in R$ for all $x \in N$. This means $x \cdot x = x^2$ must be a square of a natural number. Since $x^2$ is always a square for any natural number $x$, R is reflexive.
For R to be symmetric, if $(x, y) \in R$, then $(y, x) \in R$. If $xy$ is a square of a natural number, then $yx$ is also a square of a natural number (since multiplication is commutative). Therefore, if $(x, y) \in R$, then $(y, x) \in R$, and R is symmetric.
For R to be transitive, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$. This means if $xy$ and $yz$ are squares of natural numbers, then $xz$ must also be a square of a natural number. Let $xy = m^2$ and $yz = n^2$ for some natural numbers $m$ and $n$. Then, $x = \frac{m^2}{y}$ and $z = \frac{n^2}{y}$. So, $xz = \frac{m^2}{y} \cdot \frac{n^2}{y} = \frac{m^2n^2}{y^2} = (\frac{mn}{y})^2$. For $xz$ to be a square, $\frac{mn}{y}$ must be a natural number. Consider $x=2, y=8, z=32$. Then $xy = 2*8 = 16 = 4^2$, $yz = 8*32 = 256 = 16^2$. So $(2,8) \in R$ and $(8,32) \in R$. Now, $xz = 2*32 = 64 = 8^2$. So $(2,32) \in R$. However, consider $x=1, y=4, z=9$. $xy = 4 = 2^2$ and $yz = 36 = 6^2$. Then $xz = 9 = 3^2$. Consider $x=2, y=2, z=8$. $xy = 4 = 2^2$ and $yz = 16 = 4^2$. Then $xz = 16 = 4^2$. Consider $x=a^2, y=1, z=b^2$. $xy = a^2$ and $yz = b^2$. Then $xz = a^2b^2 = (ab)^2$. Since $xy = m^2$ and $yz = n^2$, we have $x = m^2/y$ and $z = n^2/y$. $xz = (mn/y)^2$. For $xz$ to be a perfect square, $mn/y$ must be an integer. Multiplying the two equations, we get $xy^2z = m^2n^2$, so $y^2 = \frac{m^2n^2}{xz}$. $y = \frac{mn}{\sqrt{xz}}$. If $xz$ is a perfect square, then $\sqrt{xz}$ is an integer. So, $xz$ must be a square of a natural number. Therefore, R is transitive.
Since R is reflexive, symmetric, and transitive, R is an equivalence relation.
Final Answer: R is an equivalence relation.