= Solution
Take $\lambda>0$ as the rate of the <exponential distribution>. By <independence>, the <joint probability density> of the original pair is $\lambda^2e^{-\lambda(x_1+x_2)}$ on $x_1,x_2>0$. Put $s=Y_1$, $r=Y_2$. The inverse <change of variables> is
$$
x_1=\frac{sr}{1+r},\qquad x_2=\frac{s}{1+r},\qquad s>0,\ r>0.
$$
Its <Jacobian determinant> is
$$
\det\frac{\partial(x_1,x_2)}{\partial(s,r)}
=\det\begin{pmatrix}r/(1+r)&s/(1+r)^2\\1/(1+r)&-s/(1+r)^2\end{pmatrix}
=-\frac{s}{(1+r)^2}.
$$
Using the absolute value in the <change of variables formula>, the transformed <joint probability density> is
$$
\boxed{f_{Y_1,Y_2}(s,r)=\frac{\lambda^2s e^{-\lambda s}}{(1+r)^2}\quad(s>0,r>0),}
$$
and zero otherwise. The support is the entire positive quadrant, and the factors are separately normalized:
$$
\int_0^\infty\lambda^2s e^{-\lambda s}\,ds=1,
\qquad \int_0^\infty\frac{dr}{(1+r)^2}=1.
$$
Integrating out either variable therefore yields these factors as the <marginal densities>. Factorization of the <joint probability density> proves <independence>, with
$$
\boxed{f_{Y_1}(s)=\lambda^2s e^{-\lambda s}\ (s>0),\qquad f_{Y_2}(r)=\frac1{(1+r)^2}\ (r>0).}
$$
The sum has a shape-two <gamma distribution>, and the ratio has a <beta-prime distribution> with parameters one and one, independent of $\lambda$. Its <cumulative distribution function> is $r/(1+r)$ for $r>0$.
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