I doubt that this is an inequality

If $x$ is even then $y^2 = kxy - x^2 - 10$ must be even, so $y$ is even, so $10 = kxy - x^2 - y^2$ is a multiple of $4$ . This is not true, and hence $x$ is odd. Similarly, $y$ is odd. Thus $x^2 \equiv y^2 \equiv 1 \pmod{8}$ , and hence $kxy = x^2 + y^2 + 10 \equiv 4 \pmod{8}$ . This means that $k$ is a multiple of $4$ , but not a multiple of $8$ .

If $k=4$ we would have $0 = x^2 + y^2 - 4xy + 10 \; = \; (x - 2y)^2 + 10 - 3y^2$ , so that $(x-2y)^2 + 10 = 3y^2 \equiv 0 \pmod{3}$ , so that $(x-2y)^2 \equiv 2 \pmod{3}$ . But all squares are congruent to either $0$ or $1$ modulo $3$ , so we deduce that $k \neq 4$ .

Thus the smallest possible value of $k$ is $12$ . This value of $k$ can be achieved with $x=y=1$ , and so the answer is $\boxed{12}$ .