There is nothing wrong with your answer and it does not constradicts the question.
Note when $\arcsin (V)$ in part 2, is not the same $\Theta$ YOU used in the integral.
Note $Y=y, X=x$ and $Y=-y, X=x$ gives the same value of $\arcsin(V)$, i.e.
$\arcsin(x/\sqrt{x^2+y^2}) = \arcsin(x/\sqrt{x^2+(-y)^2})$ but $X=x$ and $Y=y$ and $Y=y, X=-x$ gives different value of $\Theta$ in your parametrisation.
e.g.
$y = 1, x = 1$ gives $\Theta = \pi/4$ but $y=-1, x=1$ gives $\Theta = \pi/4 $
Explicitly
$\arcsin (V) = \Theta$ for $\Theta \in [0,\pi/2]$
$\arcsin (V) = -\Theta + \pi$ for $\Theta \in [3\pi/2, 2\pi]$
$\arcsin (V) = \Theta - \pi$ for $\Theta \in (\pi/2, \pi)$
$\arcsin (V) = -\Theta +2\pi $ for $\Theta \in [3\pi/2, 2\pi)$
so $\arcsin (V)$ is a uniform distribution on $(-\pi/2, \pi/2)$
I am physicist and I can prove it faster (as a physicist ) intuitive derivation by using https://en.wikipedia.org/wiki/Inverse_transform_sampling
Imagine $2D$ Gaussian copula in polar coordinates $(\phi,r)$ representing
$2D$ probability density in plane. (The copula is symmetric around the $Z-$axis.)
$\rho(r)=A\exp(-\frac12r^2))$ where $A$ is a scaling parameter, such that $$\int_ 0^\infty 2\pi r \rho(r) dr = 1.$$(you can easily calculate the integral ; $2\pi r dr$ is area between two concentric close circles in $x,y$ plane) You get it from linear Gauss density function in Cartesian coordinates: $\rho(r)=\rho_1(x) \rho_1(y)$ where $\rho_1(x)=\sqrt{A}\exp(-\frac12x^2))$ (using Pythagoras theorem or euclidean metric as well)
Let's generate random $3D$ points uniformly under the the Gauss copula. Projecting the uniformly distributed samples in volume $(X,Y,Z)$ or $(r,\phi,z)$ into the base plane $(X,Y)$ or $(r,\phi)$ provides the Gaussian distributed samples in the $2D$ plane. Projecting those samples into any direction (for example $X$ or $Y$ axis) we get the chosen Gaussian distribution.
Think of the following step or mathematical construction. Split the whole volume of unit 1 under the Gaussian copula into small objects of the same volume of $dV=dr .(r*d\phi) dh$ and attach an index label to them from $1$ into $N$, where $N$ is around $\frac{1}{dV}$, $dr$ is radial size,$r*d\phi$ is tangential size (perpendicular to $dr$) and $dh$ is height in $Z$ axis.Note that for the constant volume the $r*d\phi$ has to be constant, the further from origin the larger r and the smaller $d\phi$ gets to keep this tangential dimension constant.
Now you can generate uniformly integer numbers in range $i=1 \ldots N$ (or real numbers from interval $<0,1>$ multiplied by $N$ to pick the uniformly distributed index representing the uniformly distributed location of infinitesimal volume $dV_i$. This is a random uniform volume sampling. For the purists to get the true random points sampling you can combine it with random triplet $(r_1,\rho_1,h1)$ to pick the random point within the random infinitesimal volume $dV_i$. We demonstrated that uniform volume sampling leads to the Gaussian distribution of the projected samples. We choose any indexing system of infinitesimal volumes $dV$ with the only constraint: Let $dV_i$ has $r_i$ coordinate and $dV_j$ has $r_j$ coordinate. If $r_i<r_j$ then $i<j$. Indexing is growing with $r$. We leave the indexing within the wall of cylinder (volumes with the same R) unspecified on purpose.
We will generate uniformly a real number $P$ from $(0,1)$ and when multiplied by $N$ that will pick a volume $dV_i$ somewhere inside a cylinder wall (of $dr$ thickness) with radius $R$ and height $\rho(R)$ .
Probability or fraction $P$ of uniformly distributed random points under copula with $r$ closer than $R$ (inside the cylinder) is integral from $0$ to $R$ of the height $\rho(r)$ of copula at distance r from origin times thickness dr times circumference of cylinder ($2\pi r$). It is integral of this value
$P(r<R)=\int_0^R 2 \pi r \rho(r) dr$.
We can get easily (this is the reason why we work in polar coordinates) a closed formula of this integral by simple substitution $\frac12 r^2=p$ and $rdr=dp$
\begin{align}
P&=2\pi A \int_0^{\frac12 R^2} \exp(-p) dp \\
&=2\pi A \left[-\exp(-p)\right]_0^{\frac12 R^2} \\
&=2\pi A [1-\exp(-\frac12 R^2)]
\end{align}
Notice that when $R$ goes to $\infty$ we normalize the probability the integral goes to $1$ so we have $A=1/(2 \pi)$ ).
We solve for $R$:
$$R(P)=\sqrt{-2\ln(1-P)} $$
That is:An uniformly generated $P$ produces a properly distributed $R$ in polar coordinates. This is a general principle of inverse transform sampling.https://en.wikipedia.org/wiki/Inverse_transform_sampling
You generate uniform samples in generalized volume or area under a distribution function and get location (projection) of sample.
We know we should get uniformly distributed $\phi$ coordinate as well. Index $i$ generated above under fully defined specific indexing system (within cylindrical wall) should provide this uniformly distributed $\phi$ (due to symmetry) as well. However it is easier to neglect the details of indexing system within the cylinder wall and take advantage of symmetry of cupola around Z-axis. Due to rotational symmetry around Z-axis we can simply generate uniformly distributed $\phi \in (0,2\pi)$ . The pair
(R,$\phi$)=(R(P),2$\pi$*U)
where P and U are independent uniformly random distributed real values on interval (0,1) define sample points of 2D Gaussian distribution in polar coordinates. At the end we can do simple projection/transformation of those samples into Cartesian coordinates. We get the celebrated Gaussian distribution of random samples in X and Y axis independently.
$$x=R\cos(\phi) $$
$$y=R\sin(\phi)$$
In formula for $R(P)$ above notice that if $P$ is uniformly distributed on $(0,1)$, then $1-P$ is as well. You can replace $1-P$ with $P$ in the method.
$$R(P)=\sqrt{-2\ln(P)} $$
Best Answer
Well, you have done it all.
Just note that the pdf you got $f_{X,Y}$ is the density of a bivariate normal distribution with mean $\mu=(0,0)'$ and variance covariance matrix given by
$V=\left[\array{ 1 & 0\\ 0 & 1}\right]$
Hence, since the covariance is $0$ you have that $X$ and $Y$ are independent since they are normally distributed.