${\int }_3^5{\int }_1^{\sqrt{x}}2y{e}^{-x}dydx$

Step 1 ：Given the double integral problem: ${\int }_3^5{\int }_1^{\sqrt{x}}2y{e}^{-x}dydx$. The inner integral is with respect to y and the outer integral is with respect to x. The function to be integrated is $2y{e}^{-x}$. The limits of the inner integral are from 1 to $\sqrt{x}$ and the limits of the outer integral are from 3 to 5.

Step 2 ：First, solve the inner integral, treating x as a constant. The function to be integrated becomes $2y{e}^{-x}$.

Step 3 ：The result of the inner integral is $x{e}^{-x}-e^{-x}$.

Step 4 ：Next, substitute the limits of the inner integral into the result. This gives us the function to be integrated for the outer integral.

Step 5 ：Then, solve the outer integral, treating the result of the inner integral as the function to be integrated.

Step 6 ：The result of the outer integral is $-5e^{-5}+3e^{-3}$.

Step 7 ：Finally, substitute the limits of the outer integral into the result to get the final answer.

Step 8 ：The final answer is $\boxed{{\displaystyle -5e^{-5}+3e^{-3}}}$