Để ý $(x+1)(y+1)=4xy\Rightarrow xy \ge1$
Nên ta có $\frac{3y}{x(y+1)}+\frac{3x}{y(x+1)}=\frac{3y^2(x+1)+3x^2(y+1)}{xy(x+1)(y+1)}=\frac{3xy(x+y)+3(x^2+y^2)}{4x^2y^2}$$=\frac 34\left(\frac 1x+\frac 1y \right)+\frac 34\left(\frac 1{x^2}+\frac 1{y^2} \right)$
Lại có $\frac 1{x+y} \le \frac 14 \left( \frac 1x +\frac 1y \right)$
Từ đó suy ra $P \le \frac 1x+\frac 1y-\frac 14 \left(\frac 1{x^2}+\frac 1{y^2}\right)=\frac 1x+\frac 1y-\frac 14\left(\frac 1x+\frac 1y \right)^2+\frac 1{2xy}$
Mặt khác theo bất đẳng thức $AM-GM$ thì $\frac 14 \left( \frac 1x+\frac 1y \right)^2 +1 \ge \frac 1x+\frac 1y$
Suy ra $P \le1+\frac 1{2xy} \le \frac 32$
$P_{\min}=\frac 32 \Leftrightarrow x=y=1$