Let us assume that our equation has a solution. Let $x = \frac pq$ and $y = \frac ab,$ with $\gcd(p, q) = \gcd(a, b) = 1.$ Our equation is then equivalent to $p^2ab+a^2pq+q^2ab+b^2pq=3npqab \ \ (*).$ So $ab|pq(a^2+b^2) \iff ab|pq(a+b)^2 \iff ab|pq.$ Similarly, $pq|ab.$ Thus $pq = ab.$ Then $(*) \iff a^2+b^2+q^2+p^2=3nab.$ If $3 \nmid abpq,$ then $a^2+b^2+q^2+p^2=4 \pmod 3.$ Thus $3|abpq,$ so that $3|ab$ and $3|pq.$ Now, only one of $a$ and $b$ is divisible by 3, and the same thing holds for $p, q.$ Then, $3nab=a^2+b^2+q^2+p^2=0+1+0+1 \pmod 3,$ contradiction.

$ab=cd\implies a=xy,b=zt,c=xz,d=yt$