If $X$ and $Y$ are independent $N(0,\sigma^2)$, then $X^2+Y^2$ and $X/Y$ are independent? If $X$ and $Y$ are independent, then $X^2+Y^2$ and $X/Y$ are independent?
I was solving the problem for the case that $X$ and $Y$ are independent $N(0,\sigma^2)$.
So i found that $X^2+Y^2$ is negative exponential, $X/Y$ is Cauchy distribution.
How can i prove that $X^2+Y^2$ and $X/Y$ are independent?
Also, what it will be if the case for general random variable $X$, $Y$ with  $X^2+Y^2$ and $X/Y$ are defined?
 A: Joint density of $(X,Y)$ is 
$$f_{X,Y}(x,y)=\frac{1}{2\pi\sigma^2}\exp\left[-\frac{1}{2\sigma^2}\left(x^2+y^2\right)\right]\qquad,\,(x,y)\in\mathbb R^2$$
Transforming to polar coordinates (as mentioned in comments)
$$(X,Y)\to (R,\Theta)$$ such that $$X=R\cos\Theta\quad,\quad Y=R\sin\Theta$$
Then, $$R^2=X^2+Y^2\quad,\quad \cot\Theta=\frac{X}{Y}$$
Now, $$(x,y)\in\mathbb R^2\implies r>0\,,\,0<\theta<2\pi$$
Absolute value of jacobian of transformation is $$|J|=r$$ 
So joint density of $(R,\Theta)$ is 
\begin{align}
f_{R,\Theta}(r,\theta)&=\frac{r}{2\pi\sigma^2}e^{-r^2/2\sigma^2}\,\mathbf1_{r>0\,,\,0<\theta<2\pi}
\\\\&=\frac{r}{\sigma^2}e^{-r^2/2\sigma^2}\mathbf1_{r>0}\,\frac{1}{2\pi}\mathbf1_{0<\theta<2\pi}
\end{align}
Clearly, $R$ and $\Theta$ are independently distributed.
At this point, you can conclude that $R^2$ and $\cot\Theta$ are independently distributed because they are measurable functions of $R$ and $\Theta$ respectively. Or if you want to fully derive the joint density of $(R^2,\cot\Theta)$ and then conclude independence, you would have to use another change of variables. But note that this time around, the transformation would not be one-to-one.
The distribution of $X^2+Y^2$ would be a scaled $\chi^2_2$ distribution, and that of $X/Y$, as you say, would be standard Cauchy.
As for the 'general' case, we need to know the exact joint distribution of $(X,Y)$ to say anything about independence of  $(g(X,Y),h(X,Y))$ for arbitrary function $g$ and $h$.
