jmc

Let $k=\frac{y}{x}.$ Then $y=kx,$ so $$(x-3{)}^2+(kx-3{)}^2=6.$$Expressing this as a quadratic in $x,$ we get $$(k^2+1)x^2-(6k+6)k+12=0.$$This quadratic has real roots when its discriminant is nonnegative: $$(6k+6{)}^2-4(k^2+1)(12)\ge 0.$$This simplifies to $k^2-6k+1\le 0.$ The roots of the corresponding equation $k^2-6k+1=0$ are $$3\pm 2\sqrt{2},$$so the solution to $k^2-6k+1\le 0$ is $3-2\sqrt{2}\le k\le 3+2\sqrt{2}.$

Therefore, the largest possible value of $k=\frac{y}{x}$ is $\boxed{3+2\sqrt{2}}.$