show that $X/(X+Y) $ has a cauchy distribution if $X $ and $X+Y $ are standard normal A random variable $X$ has a cauchy distribution with parameters $a$ and $b$ is the density of $X$ is $f(x\mid a,b)=\dfrac{1}{\pi b}\dfrac{1}{1+(\frac{x-a}{b})^2}$ where $-\infty <x< \infty $, $-\infty <a <\infty$, $b>0$

Suppose $X$ and $Y$ are independent standard normal random variables then show that $X/(X+Y)$ has a cauchy distribution. 

Since $X$ and $Y$ are independent standard normal random variables then the joint pdf for $(X,Y)$ is $$f_{X,Y}(x,y)=\frac{1}{2\pi}e^{-x^2/2}e^{-y^2/2}$$
I used the Jacobian method to try and find $f_{U,V}$ so I let $U=X+Y$ and $V=X/(X+Y)$ so $x=uv$ and $y=u-uv$ and then $J=u$.
So $f_{U,V}(u,v)=\dfrac{u}{2\pi}e^{-(uv)^2/2}e^{-(u-uv)^2/2}=\dfrac{u}{2\pi}e^{-(uv)^2/2}e^{-(u^2-2u^2v + u^2v^2)/2}=\dfrac{u}{2\pi}e^{-(u^2v^2)/2}e^{(-u^2/2)+(u^2v) - (u^2v^2/2)}= \dfrac{u}{2\pi}e^{-(u^2v^2)-(u^2/2)+(u^2v)} = \dfrac{u}{2\pi}e^{-u^2(v^2-v+(1/2))} $ 
Therefore $$f_{U,V}(u,v)= \dfrac{u}{2\pi}e^{-u^2(v^2-v+(1/2))} ; -\infty<u,v<\infty$$
I now want to find the $V$ marginal density and that should have a cauchy distribution. 
$f_V(v)=2\int_0^{\infty} \dfrac{u}{2\pi}e^{-u^2(v^2-v+(1/2))}du=\int_0^{\infty} \dfrac{u}{\pi}e^{-u^2(v^2-v+(1/2))}du$. If I let $s=u^2$ then $\dfrac{ds}{2u}=du$ and then integral then becomes $ \int_0^{\infty} \dfrac{1}{2\pi}e^{-s(v^2-v+(1/2))}ds = \dfrac{1}{2\pi}\bigg( \dfrac{1}{v^2-v+(1/2)} \bigg)$. 
I think I might have made a mistake somewhere because I cant see how I can rearrange this to show that $f_V$ is a cauchy distribution.

You had almost reached the end. It suffices to transform:
$$\left(\dfrac{1}{2\pi}\right)\dfrac{1}{v^2-v+\frac{1}{2}}=\left(\dfrac{1}{2\pi}\right)\dfrac{1}{(v-\frac{1}{2})^2+(\frac{1}{2})^2}=\left(\dfrac{1}{2\pi}\right)\dfrac{4}{1+\left(\frac{v-\frac{1}{2}}{\frac{1}{2}}\right)^2}$$
giving
$$\left(\dfrac{1}{\pi \frac{1}{2}}\right)\dfrac{1}{1+\left(\frac{v-\frac{1}{2}}{\frac{1}{2}}\right)^2}.$$
Thus $a=b=\frac{1}{2}.$
