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.)
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.
Conclusion: A continuous function $ f: \mathbb{R} \to \mathbb{R} $ obeys the functional equation in the OP if and only if there exists a $ (c,k) \in \mathbb{R}^{2} $ such that $$ \forall x \in \mathbb{R}: \qquad f(x) = c e^{k x^{2}}. $$
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.
