Let $y\lex$. There are not solutions with $y\le 0$. Expression had maximal means when $y=x(\sqrt 3 -1)$, it means $\frac{7(3+2\sqrt 3)}{9x} <3/7$ if $x\ge 6$. Therefore $1\le y\le x \le 5$.It give y=4 x=5 (or x=4,y=5).

notice that $x+y>0$ or else LHS will become negative . The equation can also be written as $7(x+y)^2=3(x^3+y^3)$ By Holder Ineq , $x^3+y^3\geq \frac{1}{4}(x+y)^3$ so $x+y\leq \frac{28}{3}$ so $0<x+y\leq 9$ Now by letting $s=x+y$ , $p=xy$ we obtain $7s=3s^2-9p$ so $p=\frac{s(3s-7)}{9}$ We can see that $s$ must be multiple of $9$ and the only answer is when $s=9$ so $p=20$ . And this gives us $(x,y)=(4,5),(5,4)$

For integers $x,y,$ we have $x^2-xy+y^2=\left(x-\frac{y}{2}\right)^2+\frac{3}{4}y^2\geq 0.$ However since $x^2-xy+y^2$ is a denominator, we have $x^2-xy+y^2\neq 0$, yielding $x^2-xy+y^2>0.$ Thus we can set $x+y=3k,\ x^2-xy+y^2=7k$ for some positive integers $k.$ Rewrite the second equation as $(x+y)^2-3xy=7k\Longleftrightarrow xy=\frac{28k-9k^2}{3}.$ Now $(x-y)^2=(x+y)^2-4xy=\frac{1}{3}k(28-9k)\geq 0,$ the natural numbers $k=1,2,3.$ Plugg these values into the equations, the pair of integers $(x+y,xy)$ is limited to $(x+y,xy)=(9,20).$ thus the answer is $(x,y)=(4,5),\ (5,4).$

3x^2-x(3y+7)+y(3y+7)=0 (1). x is integer. So, discriminant of (1) equality is full square. D=(3y+7)^2-4y(3y+7)=(3y+7)(7-y)=k^2, where k is any positive integer. -3k^2=(3y+7)(3y-21). z=3y-7. Then, 3k^2+z^2-196=0, or (3y-7)^2+3k^2=196. If y>7, (3y-7)^2>196, so (3y-7)^2+z^2>196. So, 1<=y<=7. By checking from y=1 to y=7, we see the solutions: (x;y)=(4;5);(5;4).

We have that $7(x+y)=3x^2+3y^2-3xy+6xy-6xy$.From there we obtain : $7(x+y)=3(x+y)^2-9xy$.Because $(x+y,xy)=1$ we have that $x+y|9$.Now we question $6$ cases and get solutions: $x=4 y=5$ and $x=5 y=4$.