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$$f(x,y)=\frac{1}{2\pi}e^{\frac{-(x^{2}+y^{2})}{2}}$$ I tried the substitution: $$ U=\frac{X-Y}{X+Y},\quad V=X+Y,\\ X=\frac{UV+V}{2},\quad Y=\frac{V-UV}{2},$$ With the absoulute value of the Jacobian:

$$|\begin{vmatrix} \frac{v}{2}&\frac{u+1}{2}\\\\ \frac{-v}{2}&\frac{1-u}{2} \end{vmatrix}|=|\frac{v}{2}\cdot \frac{1-u}{2}-\frac{-v}{2}\cdot \frac{u+1}{2}|=|\frac{v(1-u+u+1)}{2}|=v$$ The joint distribution: $$f(u,v)=\frac{1}{2\pi}e^{\frac{-((\frac{uv-v}{2})^{2}+(\frac{v-uv}{2})^{2})}{2}}v=\frac{1}{2\pi}ve^{\frac{-(uv-v)^{2}}{2}}$$ The marginal distribution: $$f(u)=\frac{1}{2\pi}\int_{0}^{\infty}ve^{\frac{-(v^{2}(u-1)^{2})}{2}}\, dv$$ New substitution: $$\frac{v^{2}(u-1)^{2}}{2}=s,\quad v\, dv=\frac{1}{(u-1)^{2}}\, ds$$

$$f(u)=\frac{1}{2\pi}\frac{1}{(u-1)^{2}}\int_{0}^{\infty}e^{-s}\,ds=\frac{1}{2\pi}\frac{1}{(u-1)^{2}}(1)=\frac{1}{2\pi(u-1)^{2}}\\ 1=c\frac{1}{2\pi}\int_{-1}^{1}\frac{1}{(u-1)^{2}}\,du=c\frac{1}{2\pi}\Big[\frac{1}{u-1}\Big]_{-1}^{1}=\text{Blows up!}$$ This obviusly is not the correct distribution, but I cannot understand where things went wrong (I suspect it is that $(X=\frac{UV+V}{2})^{2}=(Y=\frac{V-UV}{2})^{2}$. Anyone that has some ideas?

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    What's $C(0,1)$? Cauchy distribution? – 2017-01-10
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    Yes, it is the Cauchy distribution. – 2017-01-10
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    Maybe you can first show that $Y/X$ has a Cauchy distribution and then use the idea that the family of Cauchy distribution is preserved under a Mobius transform (which is $x \mapsto \frac{1-x}{1+x}$ in our case). – 2017-01-10

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You have an error in your calculation of $x^2+y^2$. When you substitute the formulas for $x$ and $y$ in terms of $u$ and $v$, you should get $$ x^2+y^2=\left(\frac12(uv+v)\right)^2+\left(\frac12(v-uv)\right)^2 =\frac12\left(v^2(u^2+1)\right). $$

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    I did the calculations again. You are right. Thank you! – 2017-01-10
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There where an error in the calculation. $$f(u,v)=\frac{1}{2\pi}e^{\frac{-((\frac{uv+v}{2})^{2}+(\frac{v-uv}{2})^{2})}{2}}v=\frac{1}{2\pi}ve^{\frac{-(uv^{2}+v^{2})}{2}}$$

The correct marginal distribution:

$$f(u)=\frac{1}{2\pi}\int_{-\infty}^{\infty}ve^{\frac{-(v^{2}(u^{2}+1)}{2}}\, dv$$

Substitution:

$$\frac{v^{2}(u^{2}+1)}{2}=s,\quad v\,dv=\frac{1}{(u^{2}+1)}\,ds$$ $c$ is a constant.

$$f(u)=\frac{c}{2\pi}\frac{1}{(u^{2}+1)}(\int_{0}^{\infty}e^{-s}\, ds+\int_{0}^{\infty}e^{-s}\,ds)=\frac{c}{\pi}\frac{1}{(u^{2}+1)}(1)$$

$$1=\frac{c}{\pi}\int_{-1}^{1}\frac{1}{(u^{2}+1)}\,du=\frac{c}{\pi}\Big[\tan^{-1}(u)\Big]_{-1}^{1}=\frac{c}{\pi}(\frac{\pi}{2})=1$$

$$c=2$$

$$U\in C(0,1)$$