Prove that for all even positive integers $n = x^2 + 3y^2$ $(x, y \in \mathbb Z)$, $\dfrac{n}{4} = x'^2 + 3y'^2$ $(x, y \in \mathbb Z)$. 
Prove that for all even positive integers $n$ that can be written in the form of $x^2 + 3y^2$ $(x, y \in \mathbb Z)$, $\dfrac{n}{4}$ can be also written in the form of $x'^2 + 3y'^2$ $(x', y' \in \mathbb Z)$.

It is not difficult to establish from the start that $4 \mid n \iff 2 \mid x + y$ and vice versa.
So $n$ either equals to $(2p)^2 + 3 \cdot (2q)^2$ or $(2p + 1)^2 + 3 \cdot (2q + 1)^2$ $(m, n \in \mathbb Z)$
$\implies \dfrac{n}{4}$ either equals to $p^2 + 3q^2$ or $p^2 + 3q^2 + p + 3q + 1$
For the first case, $\dfrac{n}{4} = x'^2 + 3y'^2$ if $x' = p$ and $y' = q$.
For the second case, $\dfrac{n}{4} = p^2 + 3q^2 + p + 3q + 1$. Since $4 \not\mid n \iff 2 \not\mid x + y$,
$\dfrac{n}{4}$ either equals to $$(2p')^2 + 3 \cdot (2q' + 1)^2 + (2p') + 3 \cdot (2q' + 1) + 1 = 4p'^2 + 12q'^2 + 2p' + 18q' + 7$$ $$ = (p' - 3q' - 2)^2 + 3(p' + q' + 1)^2$$ or $$(2p' + 1)^2 + 3 \cdot (2q')^2 + (2p' + 1) + 3 \cdot (2q') + 1 = 4p'^2 + 12q'^2 + 6p' + 6q' + 3$$ $$ = (p' - 3q')^2 + 3(p' + q' + 1)^2$$ $(p', q' \in \mathbb Z)$. I want to ask if the above solution is correct or not.

The basic method you have used is correct but complications have been introduced by using $n$ for two different variables (see line 2 of your answer). A simple answer is as follows:- 
If $n$ is even then, as in your proof, we can suppose $x$ and $y$ are both odd. Then $4$ is a factor of either $x+y$ or $x-y$ and, by changing the sign of $y$ if necessary, we can suppose $4$ is a factor of $x+y$.
Then simple algebra gives $\frac{n}{4}=\left(\frac{x-3y}{4}\right)^2+3\left(\frac{x+y}{4}\right)^2.$
