Given that $X,Y$ are independent $N(0,1)$ , show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $N(0,\frac{1}{4})$ It is given that $X,Y \overset{\text{i.i.d.}}{\sim} N(0,1)$
Show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}} \overset{\text{i.i.d.}}{\sim} N(0,\frac{1}{4})$
I was thinking of making polar transformations $X=r \cos \theta, Y=r \sin \theta$
Then I am getting stuck at ranges of $\theta$

If you transform $(X,Y)\mapsto(R,\Theta)$ where $X=R\cos\Theta,Y=R\sin\Theta$, 
you should end up with the joint density of $(R,\Theta)$ as $f_{R,\Theta}(r,\theta)=\dfrac{r}{2\pi}e^{-r^2/2}\mathbf1_{\{r>0,\,0<\theta<2\pi\}}$.
This implies $R$ and $\Theta$ are independent, where $R$ has the Rayleigh distribution and $\Theta\sim\mathcal{U}(0,2\pi)$. 
Now changing variables $(R,\Theta)\mapsto(U,V)$ such that $U=R\sin(2\Theta),V=R\cos(2\Theta)$, 
you should be able to show that $U$ and $V$ are independent $\mathcal{N}(0,1)$ variables.
Note that $U=\dfrac{2XY}{\sqrt{X^2+Y^2}}$ and $V=\dfrac{X^2-Y^2}{\sqrt{X^2+Y^2}}$ are independent, which in turn means that
$\dfrac{U}{2}=\dfrac{XY}{\sqrt{X^2+Y^2}}$ and $\dfrac{V}{2}=\dfrac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $\mathcal{N}(0,1/4)$ variables.

This is independent of the post above:
Joint density of $(X,Y)$ is $\displaystyle f_{X,Y}(x,y)=\frac{1}{2\pi}e^{-\frac{1}{2}(x^2+y^2)}\,\quad,(x,y)\in\mathbb{R^2}$
We transform $(X,Y)\mapsto(R,\Theta)\mapsto(U,V)$ where 
$x=r\cos\theta\,,y=r\sin\theta$ and $u=\frac{r}{2}\sin(2\theta)\,,v=\frac{r}{2}\cos(2\theta)$
$(x,y)\in\mathbb{R^2}\implies r>0\,, 0<\theta<2\pi\implies (u,v)\in\mathbb{R^2}$. 
Note that this transformation is not one to one.
Jacobian of the transformation is $J\left(\frac{x,y}{u,v}\right) = J\left(\frac{x,y}{r,\theta}\right)J\left(\frac{r,\theta}{u,v}\right)=J_1J_2$, say.
Also, $x^2+y^2=r^2=4(u^2+v^2)$ and $|J_1||J_2|=r\times\frac{2}{r}=2$
Now $\left(U=\frac{XY}{\sqrt{X^2+Y^2}},V=\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}\right)$ has the preimages $(X,Y)$ and $(-X,-Y)$.
Moreover, $X,Y\stackrel{\text{i.i.d.}}{\sim}\mathcal{N}(0,1)\iff -X,-Y\stackrel{\text{i.i.d.}}{\sim}\mathcal{N}(0,1)$.
Hence the joint density of $(U,V)$ is given by $$f_{U,V}(u,v)=f_{X,Y}(g_1(u,v),h_1(u,v))|J_1||J_2| + f_{X,Y}(g_2(u,v),h_2(u,v))|J_1||J_2|$$
$$=\frac{1}{2\pi}e^{-\frac{1}{2} 4(u^2+v^2)}|J_1||J_2|\times 2$$
$$=\frac{1}{\sqrt{\frac{1}{4}}\sqrt{2\pi}}\exp\left(-\frac{u^2}{2\cdot\frac{1}{4}}\right)\cdot\frac{1}{\sqrt{\frac{1}{4}}\sqrt{2\pi}}\exp\left(-\frac{v^2}{2\cdot\frac{1}{4}}\right)\quad ,(u,v)\in\mathbb{R^2}$$
(We have the multiplier $2$ in the second step due to the preimages of $(x,y)$ namely $(g_i(u,v),h_i(u,v))$ for $i=1,2$, 'contributing' equally to the joint density). 
This implies $U$ and $V$ are independent $\mathcal{N}(0,1/4)$ variables.
