Find all continuous functions $f : \mathbb{R} \to \mathbb{R}$ such that $f(x) f(y)$ is radially symmetric, i.e. $$f(x) f(y) = g \left( \sqrt{x^2 + y^2} \right)$$ for some function $g : [0, \infty) \to \mathbb{R}$.
(Note: I've come up with my own solution; I was just curious to see what others would come up with.)
On
First of all $f(0)^2=g(0)$, and by defining $c:=f(0)=\sqrt{g(0)}$ we have $$ c\cdot f(x)=g(|x|)\tag1 $$ so $g(x)$ completely determines $f(x)$. Also note that $g(x)$ must be non-negative, since $$ g(x)=f(\sqrt 2/2\cdot x)^2\tag2 $$ If $c=0$ equation $(1)$ implies $g(x)\equiv 0$, and then $f(x)^2=g(\sqrt2\cdot|x|)=0$ implies $f(x)\equiv 0$.
Now let us restrict ourselves to $c\neq 0$ and $x\geq 0$. Then we have $$ f(x)^2=g(\sqrt2\cdot x)=c\cdot f(\sqrt 2\cdot x)\tag3 $$ and it turns out that this can be generalized inductively to $$ f(x)^n=c^{n-1}\cdot f(\sqrt n\cdot x),\quad\forall n\in\mathbb N\tag4 $$ Thus whenever $x$ is increased by a factor $\sqrt n$, then $y=f(x)$ is transformed by the tranformation $y\mapsto y^n/c^{n-1}$. Whenever $x$ is decreased by dividing it by $\sqrt n$, then $y$ is transformed as $y\mapsto\sqrt[n]{c^{n-1}y}$.
Combining those two, we see that if $x$ is multiplied by a factor $\sqrt{m/n}$ for some $m,n\in\mathbb N$, we see that $y$ undergoes the transformation $$ y\mapsto c^{\frac{n-m}n}\cdot y^{\frac mn}\tag5 $$ or put in terms of $f(x)$ we have $$ f\left(\sqrt{\tfrac mn}\cdot x\right)=c^{\frac{n-m}n}\cdot f(x)^{\frac mn},\quad\forall m,n\in\mathbb N\tag6 $$ Thus if we are given the two constants $c:=f(0)$ and $d:=f(1)$ we have $f(x)$ completely determined on a dense subset of $\mathbb R^+$, namely $$ f\left(\sqrt{\tfrac mn}\right)=c^{\frac{n-m}n}\cdot d^{\frac mn},\quad\forall m,n\in\mathbb N\tag7 $$
By continuity of $f$, this completely determines $f$ on all of $\mathbb R^+$, and by closer inspection on $\mathbb R$, namely $$ f(x)=c^{1-x^2}\cdot d^{x^2}=f(0)^{1-x^2}\cdot f(1)^{x^2}\tag8 $$ or put differently by defining $a:=c^{-1}\cdot d=f(1)/f(0)$ we have $$ f(x)=c\cdot a^{x^2}\tag9 $$ and BTW $g(x)=c^2\cdot a^{x^2}$. Here we have $c\in\mathbb R\setminus\{0\}$ and by equation $(1)$ we see that $c,d$ must have the same sign, so $a\in\mathbb R^+$.
We have shown that all solutions can be described as $$ f(x)=c\cdot a^{x^2}\tag{10} $$ for some pair of constants $(c,a)\in\mathbb R\times\mathbb R^+$. Note for $c=0$ the value of $a$ does not matter. This agrees with the solution set proposed by Berrick Caleb Fillmore in the comment section to the OP. And it adds no additional solutions to the set.
Here is a solution that transforms the given functional equation into a well-known one.
If $ f $ is the zero function on $ \mathbb{R} $, then we are done.
If not, suppose that $ f $ attains a non-zero value somewhere. Evidently, $ f(0) \neq 0 $, otherwise \begin{align} \forall x \in \mathbb{R}: \qquad [f(x)]^{2} & = g \! \left( \sqrt{x^{2} + x^{2}} \right) \\ & = g \! \left( \sqrt{2 x^{2}} \right) \\ & = g \! \left( \sqrt{(\sqrt{2} x)^{2} + 0^{2}} \right) \\ & = f(\sqrt{2} x) \cdot f(0) \\ & = f(\sqrt{2} x) \cdot 0 \\ & = 0. \end{align} We claim that $ f $ does not attain the value $ 0 $ anywhere. By way of contradiction, assume the contrary. As $ f(0) \neq 0 $ and $ f $ is continuous, there exists an $ a \in \mathbb{R} \setminus \{ 0 \} $ closest to $ 0 $ such that $ f(a) = 0 $. Then \begin{align} \left[ f \! \left( \frac{a}{\sqrt{2}} \right) \right]^{2} & = g \! \left( \sqrt{ \left( \frac{a}{\sqrt{2}} \right)^{2} + \left( \frac{a}{\sqrt{2}} \right)^{2} } \right) \\ & = g \! \left( \sqrt{\frac{a^{2}}{2} + \frac{a^{2}}{2}} \right) \\ & = g \! \left( \sqrt{a^{2}} \right) \\ & = g \! \left( \sqrt{a^{2} + 0^{2}} \right) \\ & = f(a) \cdot f(0) \\ & = 0 \cdot f(0) \\ & = 0. \end{align} Hence, $ f \! \left( \dfrac{a}{\sqrt{2}} \right) = 0 $. However, $ \dfrac{a}{\sqrt{2}} $ is closer to $ 0 $ than $ a $ is, which is a contradiction. Consequently, $ f $ does not attain the value $ 0 $ anywhere, and by the Intermediate Value Theorem, it is either strictly positive or strictly negative.
If $ f $ is a strictly positive solution, then $ -f $ is a strictly negative solution. Conversely, if $ f $ is a strictly negative solution, then $ -f $ is a strictly positive solution. Therefore, once we know all of the positive solutions, we will know all of the negative solutions as well.
Without loss of generality, suppose that $ f $ is strictly positive. Then $ f $ is even because $$ \forall x \in \mathbb{R}: \qquad [f(-x)]^{2} = g \! \left( \sqrt{(-x)^{2} + (-x)^{2}} \right) = g \! \left( \sqrt{x^{2} + x^{2}} \right) = [f(x)]^{2}. $$ Define a new strictly positive function $ h: \mathbb{R}_{\geq 0} \to \mathbb{R}_{> 0} $ by $$ \forall x \in \mathbb{R}_{\geq 0}: \qquad h(x) \stackrel{\text{df}}{=} \frac{f(\sqrt{x})}{f(0)}. $$ Observe that \begin{align} \forall x,y \in \mathbb{R}_{\geq 0}: \qquad h(x) \cdot h(y) & = \frac{f(\sqrt{x})}{f(0)} \cdot \frac{f(\sqrt{y})}{f(0)} \\ & = \frac{f(\sqrt{x}) \cdot f(\sqrt{y})}{[f(0)]^{2}} \\ & = \frac{g \! \left( \sqrt{(\sqrt{x})^{2} + (\sqrt{y})^{2}} \right)} {[f(0)]^{2}} \\ & = \frac{g(\sqrt{x + y})}{[f(0)]^{2}}. \end{align} On the other hand, \begin{align} \forall x,y \in \mathbb{R}_{\geq 0}: \qquad h(x + y) & = \frac{f(\sqrt{x + y})}{f(0)} \\ & = \frac{f(\sqrt{x + y}) \cdot f(0)}{[f(0)]^{2}} \\ & = \frac{g \! \left( \sqrt{(\sqrt{x + y})^{2} + 0^{2}} \right)}{[f(0)]^{2}} \\ & = \frac{g(\sqrt{x + y})}{[f(0)]^{2}}. \end{align} Hence, $ h $ satisfies the well-known functional equation $$ \forall x,y \in \mathbb{R}_{\geq 0}: \qquad h(x) \cdot h(y) = h(x + y). $$ It follows that there exists a $ k \in \mathbb{R} $ such that $$ \forall x \in \mathbb{R}_{\geq 0}: \qquad h(x) = e^{k x}. $$ We thus find that $$ \forall x \in \mathbb{R}: \qquad f(x) = f(0) \cdot e^{k x^{2}}. $$ Therefore, if $ f $ is strictly positive, then there exist a $ c \in \mathbb{R}_{> 0} $ and a $ k \in \mathbb{R} $ such that $$ \forall x \in \mathbb{R}: \qquad f(x) = c e^{k x^{2}}. $$ Any strictly positive function $ f $ on $ \mathbb{R} $ of this form clearly satisfies the functional equation in the OP.