smc

Rewrite the first equation as $$\frac{\log y}{\log x}+\frac{\log x}{\log y}=3\frac{1}{3}$$ and this is of the form $u+\frac{1}{u}=3\frac{1}{3}$; by inspection it is easy to see that $u=3$ or $\frac{1}{3}$. Therefore $\log y=3\log x$ (or vice versa--it doesn't matter here) so $y=x^3$. Substituting this into $xy=144$, this means $x=2\sqrt{3}$ and $y=24\sqrt{3}$, so $\frac{1}{2}(x+y)=13\sqrt{3}$ which is $\boxed{13\sqrt{3}}$