We have
$$\begin{aligned} & \sqrt{\frac{xy}{x^2 + y^2 + 2z^2}} + \sqrt{\frac{yz}{y^2 + z^2 + 2x^2}} + \sqrt{\frac{zx}{z^2 + x^2 + 2y^2}} \le \\ & \sqrt{\frac{xy}{xy + yz + zx + z^2}} + \sqrt{\frac{yz}{xy + yz + zx + x^2}} + \sqrt{\frac{zx}{xy + yz + zx + y^2}} = \\ & \sqrt{\frac{xy}{(z + x)(y + z)}} + \sqrt{\frac{yz}{(x + y)(z + x)}} + \sqrt{\frac{zx}{(y + z)(x + y)}} \le \\ & \frac{\frac{x}{z + x} + \frac{y}{y + z}}{2} + \frac{\frac{y}{x + y} + \frac{z}{z + x}}{2} + \frac{\frac{z}{y + z} + \frac{x}{x + y}}{2} = \\ & \frac{\frac{x + y}{x + y} + \frac{y + z}{y + z} + \frac{z + x}{z + x}}{2} = \frac{3}{2}. \end{aligned}$$
Equality holds if and only if $x = y = z$.