jmc

By the Cauchy-Schwarz inequality, $$((x+2)+2(y+2))\left(\frac{1}{x+2}+\frac{1}{y+2}\right)\ge (1+\sqrt{2}{)}^2.$$Then $$x+2+2y+4\ge 3(1+\sqrt{2}{)}^2=9+6\sqrt{2},$$so $x+2y\ge 3+6\sqrt{2}.$

Equality occurs when $(x+2{)}^2=2(y+2{)}^2,$ or $x+2=(y+2)\sqrt{2}.$ Substituting into $\frac{1}{x+2}+\frac{1}{y+2}=\frac{1}{3},$ we get $$\frac{1}{(y+2)\sqrt{2}}+\frac{1}{y+2}=\frac{1}{3}.$$Solving, we find $y=\frac{2+3\sqrt{2}}{2}.$ Then $x=1+3\sqrt{2}.$

Hence, the minimum value we seek is $\boxed{3+6\sqrt{2}}.$