Note that for $x = \dfrac{2}{\sqrt{10}}, y = \dfrac{1}{\sqrt{2}}, z = \dfrac{1}{\sqrt{10}}$ we have $|2xy + yz| = \dfrac{\sqrt{5}}{2}$ so $\lambda \le 2$. Now for $\lambda = 2$ we have $|2xy + yz| \le |2xy| + |yz| \le \dfrac{\sqrt{5}x^2 + \dfrac{4\sqrt{5}}{5}y^2}{2} + \dfrac{\dfrac{\sqrt{5}}{5}y^2 + \sqrt{5}z^2}{2} = 1$, so $\lambda = 2$ is the maximal value.