This can be rewritten as $x^2y^2+(x^2-6)^2 \leq 6$. It is apparent that $x^2y^2 \geq 0$ and $(x^2-6)^2 \geq 0$. Therefore $x=0$ is impossible. Therefore, as $x$ is an integer, we have $x^2 \geq 1$, or $x^2y^2 \geq y^2$. But $6 \geq x^2y^2 \geq y^2$, thus $y=0,\pm 1, \pm 2$. If $y=0$, we require $(x^2-6)^2 \leq 6 \implies -\sqrt{6} \leq x^2-6 \leq \sqrt{6} \implies 6-\sqrt{6} \leq x^2 \leq 6+sqrt{6}$. As $x$ is an integer, this gives $x^2=4 \implies x=\pm 2$. If $y=\pm 1$, we require $x^2+(x^2-6)^2 \leq 6 \implies x^4-11x^2+30 \leq 0 \implies (x^2-6)(x^2-5) \leq 0$. Thus we require $x^2-5 \geq 0, x^2-6 \leq 0 \implies 5 \leq x^2 \leq 6$, impossible over the integers. If $y=\pm 2$, we require $4x^2+(x^2-6)^2 \leq 6 \implies x^4-8x^2+30 \leq 0$. But $x^4+16 \geq 8x^2$ by AM-GM, so $x^4-8x^2+30 \geq 14>0$, impossible. Thus our only solutions are $(-2,0), (2,0)$.

Clearly, if $y^2\ge 12$, there is no solution. So, with $y$ being an integer, we need to check only the cases $y^2=0, 1, 4, 9$. It is easy too show that $y=0$ gives $x=\pm 2$ and that no other solution exists in the integers.

$0\geq(x^4-12x^2+x^2y^2+30)\geq(x^4-11x^2+30) $Answer: x=2 y=0x=-2 y=0