Solving the functional equation $f(x)f(y)=c\,f(\sqrt{x^{2}+y^{2}})$ 
Find all probability density function $f:\mathbb{R}\to\mathbb{R}$ such that there exists a constant $c\in\mathbb{R}$ for which $$f(x)f(y)=c\,f(\sqrt{x^{2}+y^{2}})\text{ for all }x,y\in\mathbb{R}\,.$$

The following is part of a derivation of the Gaussian distribution, from Jaynes' book "Probability theory - The logic of science" (p. 201)

Where the $f$ in question is a probability density function
The fact that $\log(\frac{f(x)}{f(0)}) = ax^2$ is a solution is obvious, but why is it the only possible one?
 A: Let $f:\mathbb{R}\to\mathbb{R}$ be a Lebesgue-measurable function such that there exists $g:\mathbb{R}_{\geq 0}\to\mathbb{R}$ for which
$$f(x)\,f(y)=g\left(\sqrt{x^2+y^2}\right)\tag{*}$$
for all $x,y\in\mathbb{R}$.  Then, note that by plugging in $y:=0$, we have
$$f(x)\,f(0)=g\big(|x|\big)\text{ for every }x\in\mathbb{R}\,.\tag{#}$$
If $f(0)=0$, then $g\equiv 0$, whence $f(x)\,f(y)=0$ for every $x,y\in\mathbb{R}$.  This shows that $f\equiv 0$ (which is not a probability density function).
We now assume that $c:=f(0)$ is nonzero.  From (*) and (#),
$$f(x)\,f(y)=c\,f\left(\sqrt{x^2+y^2}\right)\text{ for all }x,y\in\mathbb{R}\,.$$
Note that, by (#), $f$ is an even function.  Hence, it suffices to solve for $f(x)$ when $x\geq 0$.  Define $h:\mathbb{R}_{\geq 0}\to\mathbb{R}$ to be $h(t):=\dfrac{1}{c}\,f(\sqrt{t})$ for all $t\geq 0$.  Then,
$$h(s)\,h(t)=h(s+t)\text{ for all }s,t\geq 0\,.$$
That is,
$$h(t)=h\left(\frac{t}{2}+\frac{t}{2}\right)=\left(h\left(\frac{t}{2}\right)^{\vphantom{a^a}}\right)^2\geq 0\,.$$
If $h(\tau)=0$ has a solution $\tau\geq 0$, then note that
$$h(s+\tau)=h(s)\,h(\tau)=0\text{ for all }s\geq 0\,,$$
whence $h(t)=0$ for all $t\geq \tau$.
If $h(t)=0$ for all $t\geq 0$, then $f(x)=0$ for all $x\geq 0$.  This contradicts the assumption that $f(0)\neq 0$.  If there exists $t\geq 0$ such that $h(t)\neq 0$, then let $$\sigma:=\sup\big\{t\geq 0\,\big|\,h(t)=0\big\}\,.$$
Note that $\sigma\leq \tau$.  If $\sigma>0$, then note that
$$0=h(2\sigma)=\big(h(\sigma)\big)^2\,,$$
whence $h(\sigma)=0$.  That is,
$$0=h(\sigma)=\left(h\left(\frac{\sigma}{2}\right)^{\vphantom{a^a}}\right)^2\,,$$
implying $h\left(\dfrac{\sigma}{2}\right)=0$.  By the same argument as the previous paragraph, $h(t)=0$ for all $t\geq \dfrac{\sigma}{2}$.  This contradicts the definition of $\sigma$.  Thus, $\sigma=0$.  Now,
$$h(0)=h(0+0)=\big(h(0)\big)^2$$
implies that $h(0)=0$ or $h(0)=1$.  Since $\sigma=0$, we get $h(0)=1$, leading to a solution
$$h(t)=\left\{\begin{array}{ll}1&\text{if }t=0\,,\\0&\text{if }t>0\,.\end{array}\right.$$
That is,
$$f(x)=\left\{\begin{array}{ll}c&\text{if }x=0\,,\\0&\text{if }x\neq 0\,.\end{array}\right.$$
(This makes $f$ a measurable function, but this solution is not a probability density function.)
From now on, we assume that $h(t)>0$ for all $t\geq 0$.  Define
$$\eta(t):=\ln\big(h(t)\big)\text{ for }t\ge 0\,.$$
Then,
$$\eta(s+t)=\eta(s)+\eta(t)\text{ for all }s,t\geq 0\,.\tag{@}$$
This is Cauchy's functional equation.  Since $f$ is measurable, $h$ is also measurable, and so is $\eta$.  Therefore, there exists a constant $a\in\mathbb{R}$ for which
$$\eta(t)=at\text{ for each }t\geq 0\,.$$
That is,
$$h(t)=\exp(at)\text{ for each }t\geq 0\,,$$
whence
$$f(x)=c\,\exp(ax^2)\text{ for all }x\in\mathbb{R}\,.$$
Now, if $f$ is a probability density function, $a=-b$ for some $b>0$, and $c=\sqrt{\dfrac{b}{\pi}}$.
Remark.  There are solutions $f$ that are not Lebesgue measurable.  All such solutions are given by $$f(x):=c\,\exp\big(\eta(x^2)\big)$$ for all $x\in\mathbb{R}$, where $\eta:\mathbb{R}_{\geq 0}\to\mathbb{R}$ is any solution to (@) which is not Lebesgue measurable.  To find such $\eta$, you need the Axiom of Choice.
A: $$
\log\frac{f(x)}{f(0)} + \log\frac{f(y)}{f(0)} = \log\frac{f\left(\sqrt{x^2+y^2}\,\right)}{f(0)}
$$
Differentiating both sides with respect to $x,$ we get
$$
\frac{f'(x)}{f(x)} = \frac{f'\left( \sqrt{x^2+y^2}\, \right)}{f\left( \sqrt{x^2+y^2}\,\right)} \cdot \frac x {\sqrt{x^2+y^2}}
$$
Then
$$
\frac{f'(x)}{f(x)}\cdot \frac 1 x = \frac{f'\left( \sqrt{x^2+y^2}\, \right)}{f\left( \sqrt{x^2+y^2}\,\right)} \cdot \frac 1 {\sqrt{x^2+y^2}}
$$
But the same argument with $y$ instead of $x$ shows that $\dfrac{f'(y)}{f(y)}\cdot \dfrac 1 y$ is equal to the expression on the right above. Consequently we have
$$
\frac{f'(x)}{f(x)} \cdot \frac 1 x = \frac{f'(y)}{f(y)} \cdot \frac 1 y.
$$
Since the expression on the right in this last equation does not change as $x$ changes with $y$ held fixed, neither does the one on the left. So we have
$$
\frac{f'(x)}{f(x)} \cdot \frac 1 x = \text{some constant, i.e. something not depending on } x.
$$
$$
\frac{f'(x)}{f(x)} = Cx.
$$
$$
\log f(x) = C \frac{x^2}2 + B.
$$
$$
f(x) = e^{Cx^2/2} \cdot A.
$$
