设$x$，$y$，$z$，$w$是正实数，则$\dfrac{xy+2yz+3zw}{{x}^{2}+{y}^{2}+{z}^{2}+{w}^{2}}$的最大值为                 ．

$\because x$，$y$，$z$，$w$是正实数，

$\therefore \dfrac{xy+2yz+3zw}{{x}^{2}+{y}^{2}+{z}^{2}+{w}^{2}}$

$=\dfrac{xy+2yz+3zw}{{x}^{2}+\lambda {y}^{2}+(1-\lambda ){y}^{2}+(1-\mu ){z}^{2}+\mu {z}^{2}+{w}^{2}}$

$\leqslant \dfrac{xy+2yz+3zw}{2\sqrt{\lambda }xy+2\sqrt{(1-\lambda )(1-\mu )}yz+2\sqrt{\mu }zw}$，$\lambda \in (0,1)$，

由$2\sqrt{\lambda }:2\sqrt{(1-\lambda )(1-\mu )}:2\sqrt{\mu }=1:2:3$，整理得$\mu =9\lambda$，

$\therefore \dfrac{\sqrt{\lambda }}{\sqrt{(1-\lambda )(1-9\lambda )}}=\dfrac{1}{2}$，整理得$9\lambda ^{2}-14\lambda +1=0$，

解得$\lambda =\dfrac{7-2\sqrt{10}}{9}$或$\lambda =\dfrac{7+2\sqrt{10}}{9}$（舍$)$，

$\therefore \dfrac{xy+2yz+3zw}{{x}^{2}+{y}^{2}+{z}^{2}+{w}^{2}}\leqslant \dfrac{1}{2\sqrt{\dfrac{7-2\sqrt{10}}{9}}\cdot \dfrac{xy+2yz+3zw}{xy+2yz+3zw}}=\dfrac{1}{2\sqrt{\dfrac{7-2\sqrt{10}}{9}}}=\dfrac{3}{2\sqrt{(\sqrt{5}-\sqrt{2})^{2}}}=\dfrac{\sqrt{5}+\sqrt{2}}{2}$，

$\therefore \dfrac{xy+2yz+3zw}{{x}^{2}+{y}^{2}+{z}^{2}+{w}^{2}}$的最大值为$\dfrac{\sqrt{5}+\sqrt{2}}{2}$．

故答案为：$\dfrac{\sqrt{5}+\sqrt{2}}{2}$