Is the product of two Gaussian random variables also a Gaussian?

The product of two Gaussian random variables is distributed, in general, as a linear combination of two Chi-square random variables:

$$ XY \,=\, \frac{1}{4} (X+Y)^2 - \frac{1}{4}(X-Y)^2$$

Now, $X+Y$ and $X-Y$ are Gaussian random variables, so that $(X+Y)^2$ and $(X-Y)^2$ are Chi-square distributed with 1 degree of freedom.

If $X$ and $Y$ are both zero-mean, then

$$ XY \sim c_1 Q - c_2 R$$

where $c_1=\frac{Var(X+Y)}{4}$, $c_2 = \frac{Var(X-Y)}{4}$ and $Q, R \sim \chi^2_1$ are central.

The variables $Q$ and $R$ are independent if and only if $Var(X) = Var(Y)$.

In general, $Q$ and $R$ are noncentral and dependent.

As @Yemon Choi showed in the first question, without any hypothesis the answer is negative since $P(X^2<0)=0$ whereas $P(U<0)\neq 0$ if $U$ is Gaussian.

For the second question the answer is also no. Take $X$ and $Y$ two Gaussian random variables with mean $0$ and variance $1$. Since they have the same variance, $X-Y$ and $X+Y$ are independent Gaussian random variables. Put $Z:=\frac{X^2-Y^2}2=\frac{X-Y}{\sqrt 2}\frac{X+Y}{\sqrt 2}$. Then $Z$ is the product of two independent Gaussian, but the characteristic function of $Z$ is $\varphi_Z(t)=\frac 1{\sqrt{1+t^2}}$, which is not the characteristic function of a Gaussian.

You can use moments to see that the product $XY$ of independent normals cannot be normal except in trivial cases. By trivial, I mean $\mathbb{V}(X)\mathbb{V}(Y)=0.$

Suppose that $X,Y$ are independent normals so that $XY$ is normal.

Case 1: Suppose that $\mathbb{E}(X)=0$. By independence, $\mathbb{E}(XY)=\mathbb{E}(X)\mathbb{E}(Y)=0$, so $XY$ is mean zero normal, and hence $$\mathbb{E}((XY)^4)=3\mathbb{E}((XY)^2)^2.$$ By independence we get
$$\mathbb{E}(X^4)\mathbb{E}(Y^4)=3\mathbb{E}(X^2)^2\mathbb{E}(Y^2)^2.$$ Either $\mathbb{V}(X)=0$, or dividing by $\mathbb{E}(X^4)$ gives $ \mathbb{E}(Y^4)= \mathbb{E}(Y^2)^2.$ This shows that $Y^2$, and hence $Y$, has zero variance.

Case 2: Suppose that $\mathbb{E}(X^2)>0$ and $\mathbb{E}(Y^2)>0$. Then, without loss of generality, $\mathbb{E}(X^2)=1$ and $\mathbb{E}(Y^2)=1$. In this case, we also have $\mathbb{E}((XY)^2)=1$, so $$\begin{eqnarray} \mathbb{E}(X^3)&=&\mathbb{E}(X)(3-2\mathbb{E}(X)^2)\\ \mathbb{E}(Y^3)&=&\mathbb{E}(Y)(3-2\mathbb{E}(Y)^2)\\ \mathbb{E}((XY)^3)&=&\mathbb{E}(XY)(3-2\mathbb{E}(XY)^2) \end{eqnarray} $$ Subtracting the product of the first two lines from the third line gives $$0=6\mathbb{E}(X)\,\mathbb{E}(Y)\,\mathbb{V}(X)\,\mathbb{V}(Y).$$ Either we are in a trivial case, or back to Case 1.

Thus, the product cannot be normal except in trivial cases.