Prove that these $3$ functions are constant given a relation is satisfied by divisors of them This particular question was asked in masters exam for which I am preparing and I am not able to solve it .

Question : suppose $f,g,h$ are functions from the set of positive real numbers into itself  satisfying $f(x)g(y)=h((x^2 +y^2)^{1/2})$ for all $x,y \in(0,\infty)$ . Then show that the three functions $\frac{f(x)}{g(x)} ,\frac{g(x)}{h(x)}$ and $\frac{h(x)}{f(x)}$ are all constant .


Attempt : I first tried by putting $y=0$ only to realize that $0$ is not in domain of $g(y)$ . I am unable to think about any other approach . I am sorry but cant give any other thing in attempt .
Kindly tell how should I approach this question .
 A: In this answer, I show that $f$ must be of the form  $f(x)=e^{c_1x^2+c_2}$,
where $c_1$ and $c_2$ are two constants, and your conclusion easily follows.
As noted in DanielFischer's and AlexRavsky's comments, there is a constant $c\gt 0$ such that $g(x)=cf(x)$ for $x\gt 0$, so that $h(\sqrt{x^2+y^2})=cf(x)f(y)$ for $x,y\geq 0$, or
$\ln(h(\sqrt{x^2+y^2}))=\ln(c)+\ln(f(x))+\ln(f(y))$. Putting $H(t)=\ln(h(\sqrt{t}))$ and $F(t)=\ln(f(\sqrt{t}))$ for $t\gt 0$, we deduce
$$
H(x^2+y^2)=\ln(c)+F(x^2)+F(y^2)\ (x,y\gt 0)\tag{1}
$$
or (putting $a=x^2,b=y^2$),
$$
H(a+b)=\ln(c)+F(a)+F(b) \ (a,b\gt 0) \label{2}\tag{2}
$$
Taking $b=a$ in \eqref{2}, we deduce $H(2a)=\ln(c)+F(2a)$ or $H(t)=\ln(c)+2F(\frac{t}{2})$ and hence $H(a+b)=\ln(c)+2F(\frac{a+b}{2})$ ; comparing with \eqref{2} we deduce
$$
F\bigg(\frac{a+b}{2}\bigg)=\frac{F(a)+F(b)}{2} \ (a,b\gt 0)\label{3}\tag{3}
$$
If we take  $a=x,b=x+2y$ in \eqref{3}, we deduce
$$
F(x+2y)=2F(x+y)-F(x) \ (x,y\gt 0)\label{4}\tag{4}
$$
Replacing $x$ with $x+y$ in \eqref{4}, we deduce $F(x+3y)=3F(x+y)-2F(x)$. More generally, by induction on $k\geq 0$ we have
$$
F(x+ky)=kF(x+y)-(k-1)F(x) \ (x,y\gt 0, k\in{\mathbb N})\label{5}\tag{5}
$$
Now, let $t\in {\mathbb Q}_+\cap [0,1]$, so that $t$ can be written $t=\frac{p}{q}$ where $p,q$ are integers and $q\geq 1, 0 \leq p \leq q$.
Let $a,b \geq 0$. Suppose that $a\leq b$. Putting $x=a, y=\frac{b-a}{q}$.
For $k=p$ in \eqref{5}, we see that $F((1-t)a+tb)=F(x)+p(F(x+y)-F(x))$.
For $k=q$ in \eqref{5}, we see that $F(b)=F(x)+q(F(x+y)-F(x))$.
Combining those two, we see that $F((1-t)a+tb)=(1-t)F(a)+tF(b)$. When $b\lt a$, we obtain the same equality by a similar argument (tahe $x=b,y=\frac{a-b}{q}$, $k=q-p$ or $q$). So we have just shown :
$$
F((1-t)a+tb)=(1-t)F(a)+tF(b) \ (a,b\gt 0,t\in {\mathbb Q}_+\cap [0,1])\label{6}\tag{6}
$$
Now, let $t\in {\mathbb Q}_+$ with $t\gt 1$. Then $t'=\frac{1}{t}\in (0,1)$, and if we put $a'=a, b'=(1-t)a+tb$, from \eqref{6} we deduce $F((1-t')a'+t'b')=(1-t')F(a')+t'F(b')$, or $F(b)=(1-\frac{1}{t})F(a)+F((1-t)a+tb)$. We thus see that \eqref{6} still holds when $t\in {\mathbb Q}_+$ :
$$
F((1-t)a+tb)=(1-t)F(a)+tF(b) \ (a,b\gt 0,t\in {\mathbb Q}_+)\label{6'}\tag{6'}
$$
Now, let $t\in {\mathbb Q}_-$ with $(1-t)a+tb \gt 0$. Then $|t|\in {\mathbb Q}_+$, so by (4') above we have $F((1-|t|)b+|t|a)=(1-|t|)F(b)+|t|F(a)$, so that (4') still holds when $t$ is negative (on the condition that $(1-t)a+tb \gt 0$) :
$$
F((1-t)a+tb)=(1-t)F(a)+tF(b) \ (a,b\gt 0,t\in {\mathbb Q}, (1-t)a+tb \gt 0)\label{6''}\tag{6''}
$$
But then, by the accepted answer in another MSE question, we know that $F$ is affine, i.e. $F(x)=c_1x+c_2$ where $c_1$ and $c_2$ are two constants.
Returning to $f$, one obtains $f(x)=e^{c_1x^2+c_2}$, which finishes the proof.
