Can the product of two non-independent Gaussians be Gaussian? We recently discussed this: Is the product of two Gaussian random variables also a Gaussian? What was established was that in nontrivial cases (i.e., ruling out zero-variance Gaussians, which are Dirac delta functions), the product of independent random variables with Gaussian distributions cannot have a Gaussian distribution. It's also easy to come up with examples of non-independent Gaussians whose product isn't Gaussian (e.g., the product of a Gaussian with itself). 
But can one construct a nontrivial case in which the product of two non-independent Gaussian-distributed random variables is also Gaussian? If so, is it possible to do it with a joint distribution that is a well-behaved function, or only with one that is highly discontinuous or a generalized function?
As a possible approach in constructing such an example, suppose we start with independent Gaussians $X_0$ and $Y_0$, which both have mean 0 and standard deviation 1. Then $A_0=(10+X_0)(10+Y_0)$ is approximately Gaussian. The most noticeable deviation from Gaussianity would probably be that the $A_0$'s distribution would be skewed. I would imagine that we could then make up random variables $X_1$ and $Y_1$ with mean 0 and standard deviation 1, independent of $X_0$ and $Y_0$ but not independent of each other, such that we get zero skewness for $A_1=(10+aX_0+\sqrt{1-a^2}X_1)(10+aY_0+\sqrt{1-a^2}Y_1)$. Continuing in this way, we might be able to make all the moments of $A_n$ have  Gaussian values as $n\rightarrow\infty$.
Another approach might be to take $X$ to be Gaussian, and $Y=f(X)$ for some function $f$ such that $f=f^{-1}$, so that $Y$ has the same distribution as $X$. It seems like if you had the freedom to make $f$ ill-behaved (maybe discontinuous everywhere), you ought to be able to make $XY$ have the desired distribution.
 A: Here is a partial solution.
If $(X,Y)$ is jointly normal, with $\sigma^2_x \sigma_y^2>0$ (non-trivial case)
 and $\mu_x=0$ then $XY$ cannot be normal. 
If it were, then the third cumulant of $XY$
$$\kappa_3= 2\, \sigma_x\, \sigma_y(\sigma_y^2\, \sigma_x^2\, \rho^3\, 
+3\, \sigma_y^2\, \sigma_x^2\, \rho
+3\, \sigma_x^2\, \rho\, \mu_y^2
+3\, \sigma_y\, \sigma_x\, \rho^2\, \mu_y\, \mu_x
+3\, \sigma_y\, \sigma_x\, \mu_x\, \mu_y
+3\, \sigma_y^2\rho\, \mu_x^2)$$
would vanish. Since $\mu_x=0$ this leads to 
$$0=\rho(\rho^2 \sigma_y^2+3\sigma_y^2+3\mu_y^2).$$
Then $\rho=0$ and we are back in the independent case. 

Update: 
The more I think about it, the more I realize
that the result is not even plausible.
Generally speaking, the product $XY$ is much too likely to take values near zero
to be normal, or any other nice distribution. 
Suppose that  $(X,Y)$ has any "nice" bivariate distribution, with 
a joint density function that is continuous and non-zero at the origin.
We argue that $XY$ cannot have a "nice" density at zero.
For $0<u<1$ define $$A_u=\{(x,y):u^{1/2}<x<u^{1/4},\, 0<y<u/x\}.$$
These sets shrink towards the origin as $u\to 0$. 
We have
$$\mathbb{P}(0<XY\leq u)\geq \mathbb{P}((X,Y)\in A_u),$$
and so letting $\lambda$ denote two-dimensional Lebesgue measure
also 
$${\mathbb{P}(0<XY\leq u)\over u}
\geq {\mathbb{P}((X,Y)\in A_u)\over\lambda(A_u)}\cdot{\lambda(A_u)\over u} 
= {\mathbb{P}((X,Y)\in A_u)\over\lambda(A_u)}\cdot{\log(1/u)\over 4}.$$
As $u\to 0$, the first factor on the right converges to $f_{(X,Y)}(0,0)$ and the
second one blows up. Therefore, the left hand side also blows up, showing that  $XY$ 
cannot have a continuous density with finite value at $u=0$.   
