Obviously $X/(X+Y)$ is between $0$ and $1$. Let (lower-case) $w$ be a number between $0$ and $1$.

$$ \frac{X}{X+Y} \le w \text{ if and only if } (1-w)X \le wY. $$ We will find this probability.

The conditional probability that $Y\ge\dfrac{(1-w)}{w}X$, given the value of $X$, is $$ e^{-(1-w)X/w}. $$

The probability we seek is then the expected value of that: \begin{align} \mathbb E e^{-(1-w)X/w} & = \int_0^\infty e^{-(1-w)x/w} \Big(e^{-x}\,dx\Big) \\[10pt] & = \int_0^\infty e^{-x/w} \, dx \\[10pt] & = w. \end{align}

In other words, this random variable is uniformly distributed between $0$ and $1$.

Second method: The equality $Y=\frac{1-w}{w}X$ is a straight line through the $(X,Y)$-plane, passing through $(0,0)$ and having positive slope.

We can let $y$ go from $0$ to $\infty$ and then for any fixed value of $y$, let $x$ go from $0$ to $\frac{w}{1-w}y$ \begin{align} & {}\qquad\int_0^\infty \left(\int_0^{wy/(1-w)} e^{-y} e^{-x} \,dx \right) \, dy \\[10pt] & = \int_0^\infty e^{-y}\left(1-e^{-wy/(1-w)}\right) \, dy \\[10pt] & = \int_0^\infty e^{-y} - e^{-y/(1-w)} \, dy \\[10pt] & = 1 - (1-w) \\[10pt] & = w. \end{align}

Obviously $X/(X+Y)$ is between $0$ and $1$. Let (lower-case) $w$ be a number between $0$ and $1$.

$$ \frac{X}{X+Y} \le w \text{ if and only if } (1-w)X \le wY. $$ We will find this probability.

The conditional probability that $Y\ge\dfrac{(1-w)}{w}X$, given the value of $X$, is $$ e^{-(1-w)X/w}. $$

The probability we seek is then the expected value of that: \begin{align} \operatorname{E} e^{-(1-w)X/w} & = \int_0^\infty e^{-(1-w)x/w} \Big(e^{-x}\,dx\Big) \\[10pt] & = \int_0^\infty e^{-x/w} \, dx \\[10pt] & = w. \end{align}

In other words, this random variable is uniformly distributed between $0$ and $1$.

Second method: The equality $Y=\frac{1-w}{w}X$ is a straight line through the $(X,Y)$-plane, passing through $(0,0)$ and having positive slope.

We can let $y$ go from $0$ to $\infty$ and then for any fixed value of $y$, let $x$ go from $0$ to $\frac{w}{1-w}y$ \begin{align} & \int_0^\infty \left(\int_0^{wy/(1-w)} e^{-y} e^{-x} \,dx \right) \, dy \\[10pt] = {} & \int_0^\infty e^{-y}\left(1-e^{-wy/(1-w)}\right) \, dy \\[10pt] = {} & \int_0^\infty e^{-y} - e^{-y/(1-w)} \, dy \\[10pt] = {} & 1 - (1-w) \\[10pt] = {} & w. \end{align}

Obviously $X/(X+Y)$ is between $0$ and $1$. Let (lower-case) $x$ be a number between $0$ and $1$.

$$ \frac{X}{X+Y} \le x \text{ if and only if } (1-x)X \le xY. $$ We will find this probability.

The conditional probability that $Y\ge\dfrac{(1-x)}{x}X$, given the value of $X$, is $$ e^{-(1-x)X/x}. $$

The probability we seek is then the expected value of that: \begin{align} \mathbb E e^{-(1-x)X/x} & = \int_0^\infty e^{-(1-x)w/x} \Big(e^{-w}\,dw\Big) \\[10pt] & = \int_0^\infty e^{-w/x} \, dw \\[10pt] & = x. \end{align}

In other words, this random variable is uniformly distributed between $0$ and $1$.

Obviously $X/(X+Y)$ is between $0$ and $1$. Let (lower-case) $w$ be a number between $0$ and $1$.

$$ \frac{X}{X+Y} \le w \text{ if and only if } (1-w)X \le wY. $$ We will find this probability.

The conditional probability that $Y\ge\dfrac{(1-w)}{w}X$, given the value of $X$, is $$ e^{-(1-w)X/w}. $$

The probability we seek is then the expected value of that: \begin{align} \mathbb E e^{-(1-w)X/w} & = \int_0^\infty e^{-(1-w)x/w} \Big(e^{-x}\,dx\Big) \\[10pt] & = \int_0^\infty e^{-x/w} \, dx \\[10pt] & = w. \end{align}

In other words, this random variable is uniformly distributed between $0$ and $1$.

Second method: The equality $Y=\frac{1-w}{w}X$ is a straight line through the $(X,Y)$-plane, passing through $(0,0)$ and having positive slope.

We can let $y$ go from $0$ to $\infty$ and then for any fixed value of $y$, let $x$ go from $0$ to $\frac{w}{1-w}y$ \begin{align} & {}\qquad\int_0^\infty \left(\int_0^{wy/(1-w)} e^{-y} e^{-x} \,dx \right) \, dy \\[10pt] & = \int_0^\infty e^{-y}\left(1-e^{-wy/(1-w)}\right) \, dy \\[10pt] & = \int_0^\infty e^{-y} - e^{-y/(1-w)} \, dy \\[10pt] & = 1 - (1-w) \\[10pt] & = w. \end{align}