First, we claim $xz\ne 1$. This is because if $xz=1$, then we will have $x+y=0$ which contradicts the condition that $x$ and $z$ are both positive. Hence, we can conclude $$xyz+x+z=y\Leftrightarrow y=\frac{x+z}{1-xz}$$ Let $x=\tan \alpha$, $y=\tan\beta$ and $z=\tan\gamma$ where $\alpha, \beta, \gamma\in (0, 90^\circ)$. $$\tan\beta = \frac{\tan\alpha + \tan\gamma}{1-\tan\alpha\tan\gamma}=\tan(\alpha+\gamma)\implies \beta = \alpha+\gamma$$ Then, we have \begin{align*} P&= \frac{2}{\tan^2\alpha + 1}-\frac{2}{\tan^2(\alpha + \gamma)+1} + \frac{3}{\tan^2\gamma + 1}\\ &= 2\cos^2\alpha -2\cos^2(\alpha + \gamma) + 3\cos^2\gamma\\ &= (1+\cos 2\alpha) - (1+\cos 2(\alpha+\gamma)) + 3\cos^2\gamma\\ &= (\cos 2\alpha - \cos 2(\alpha+\gamma)) + 3\cos^2\gamma \\ &= 2\sin\alpha\sin(2\alpha + \gamma) +3(1-\sin^2\gamma)\\ &\le 2\sin\gamma + 3 - 3\sin^2\gamma\\ &= -3\Big(\sin\gamma -\frac{1}{3}\Big)^2 +\frac{10}{3}\\ &\le\boxed{\frac{10}{3}} \end{align*} Equality holds if and only if $$ \left\{ \begin{array}{rl} \sin(2\alpha+\gamma) &= 1\\ \sin\gamma &=\frac{1}{3} \end{array} \right. \implies \left\{ \begin{array}{rl} \sin\gamma &=\frac{1}{3}\\ 2\alpha +\gamma&=\frac{\pi}{2} \end{array} \right. \implies \left\{ \begin{array}{rl} x &=\frac{\sqrt{2}}{2}\\ y &=\sqrt{2}\\ z &=\frac{\sqrt{2}}{4} \end{array} \right. $$