Đặt $\frac{6-xy}{x+y}=z\Leftrightarrow xy+yz+zx=6$ Khi đó $P=8x^2+13y^2+z^2$
Ta có $10x^2+15y^2+3z^2=\frac{x^2}{\dfrac 1{10}}+\frac{y^2}{\dfrac 1{15}}+\frac{z^2}{\dfrac 13} \ge \frac{(x+y+z)^2}{\dfrac 1{10}+\dfrac 1{15}+\dfrac13}=2(x+y+z)^2$
$\Leftrightarrow 8x^2+13y^2+z^2 \ge 4(xy+yz+zx)=24$
$\Rightarrow \min P=24$ đạt đc tại $x=\sqrt{\frac{27}{28}},y=\sqrt{\frac 37}$