$\begin{aligned} & x+i y=\sqrt{\frac{3+i}{1+3 i}}=\sqrt{\frac{(3+i)}{(1+3 i)} \times \frac{(1-3 i)}{(1-3 i)}} \\ &= \sqrt{\frac{6-8 i}{10}}=\sqrt{\frac{3}{5}-\frac{4}{5} i} \\ & \therefore \quad(x+i y)^2=\frac{3}{5}-\frac{4}{5} i \\ & \Rightarrow \quad x^2-y^2+2 x y i=\frac{3}{5}-\frac{4}{5} i \\ & \Rightarrow \quad x^2-y^2=\frac{3}{5}, 2 x y=\frac{-4}{5} \\ & \Rightarrow \quad\left(x^2+y^2\right)^2=\left(x^2-y^2\right)^2+4 x^2 y^2 \\ &=\left(\frac{3}{5}\right)^2+\left(\frac{-4}{5}\right)^2=\frac{9}{25}+\frac{16}{25}=1 .\end{aligned}$