jmc

Clearly $x+y$ is the largest of $x,x+y,x-y$, and this value is thus the length of the triangle's hypotenuse. By the Pythagorean theorem, we see that $x^2+{\left(x-y\right)}^2={\left(x+y\right)}^2$. Dividing through by $x^2$, we find that $1+{\left(1-\frac{y}{x}\right)}^2={\left(1+\frac{y}{x}\right)}^2$.

Rearranging and factoring a difference of squares, we obtain $1=2\left(2\cdot \frac{y}{x}\right)$. Thus we see that $y÷x=\boxed{\frac{1}{4}}$.