Let the joint distribution of $X$ and $Y$ be $f(x,y)=2e^{-(x+y)}$ if $0<x<y$. Without integration, only considering the exponential order statistics show that $P(Y<3X)=1/2$.
From the joint distribution of order statistics I deduced that $f(x)=e^{-x}$ and $f(y)=e^{-y}$. But I don't understand how to show the rest without integrating. Could you help please?
Joint density of $(X,Y)$ is $$f(x,y)=2e^{-(x+y)}\mathbf1_{0<x<y}=e^{-(y-x)}\mathbf1_{y>x}\,2e^{-2x}\mathbf1_{x>0}\tag{1}$$
You are right that this is the joint density of the minimum and maximum order statistics for a sample of size 2 drawn from $\mathsf{Exp}(1)$ population.
What you can also see from $(1)$ is that $Z=Y-X\sim\mathsf{Exp}(1)$, independent of $X$.
(Here is a relevant post of the above fact.)
Therefore, without any actual integration,
\begin{align} P(Y<3X)&=P(Z<2X) \\&=\int P(Z<2x\mid X=x)f_{X}(x)\,dx \\&=2\int_0^\infty (1-e^{-2x})e^{-2x}\,dx \\&={\int_0^\infty 2e^{-2x}\,dx}-\frac{1}{2}\int_0^\infty 4e^{-4x}\,dx \\&=1-\frac{1}{2}=\frac{1}{2} \end{align}
While this might not be what you are looking for, this is one possible way.
