Functions $f(x)/g(x), g(x)/h(x),h(x)/f(x)$ are constant Suppose $f,g,h$ are functions from the set of positive real numbers into itself satisfying $f(x)g(y)=h(\sqrt{x^2+y^2})$ for all $x,y\in (0,\infty)$. 
Show that the functions  $f(x)/g(x), g(x)/h(x),h(x)/f(x)$ are constant.
I guess this is equivalent to prove that their derivatives are zero.
$\left(\frac{f(x)}{g(x)}\right)'=\frac{g(x)f'(x)-f(x)g'(x)}{\left(g(x)^2\right)}=0$ if and only if $g(x)f'(x)-f(x)g'(x)=0$ i.e., $g(x)f'(x)=f(x)g'(x)$.
For similar reasons, we must have $g(x)h'(x)=h(x)g'(x)$ and $h(x)f'(x)=f(x)h'(x)$ for all $x$...
I was trying to consider $f(x)g(x)=h(x\sqrt{2})$ which would imply $f(x)g'(x)+g(x)f'(x)=\sqrt{2}h'(x\sqrt{2})$.
I am lost from here...
Help me to solve this...
I was thinking of taking the step from first principles..
We have $$\left(\frac{f(x)}{g(x)}\right)'=\lim_{t\rightarrow 0}\frac{1}{t}\left(\frac{f(x+t)}{g(x+t)}-\frac{f(x)}{g(x)}\right)=\lim_{t\rightarrow 0}\frac{1}{t}\left(\frac{f(x+t)g(x)-g(x+t)f(x)}{g(x+t)g(t)}\right)=\lim_{t\rightarrow 0}\frac{1}{t}\left(\frac{h(\sqrt{(x+t)^2+x^2}-h(\sqrt{(x+t)^2+x^2}}{g(x+t)g(t)}\right)=0$$
So, $f(x)/g(x)$ is constant function..
 A: EDIT: I am using continuity of the function $h$ below. (Which is not among the assumptions in the question.) I do not know how to proceed without this assumption. (Unless I missed something, continuity of $h$ at one point of the domain is sufficient for this proof to go through.)

Notice that for any $x,y\in(0,\infty)$ you have
$$f(x)g(y)=h(\sqrt{x^2+y^2})=f(y)g(x)$$
which implies
$$\frac{f(x)}{g(x)}=\frac{f(y)}{g(y)}.$$
This implies that $f(x)/g(x)$ is a constant function.
As far as $h(x)/f(x)$ is concerned, we get for any $r>0$ and $\varphi\in(0,\frac\pi2)$:
$$h(r)=f(r\cos\varphi)g(r\sin\varphi)$$
which implies (for any fixed $r>0$, taking limit $\varphi\to0$)
$$h(r)=f(r) \lim\limits_{t\to0^+} g(t),$$
assuming the limit $\lim\limits_{t\to0^+} g(t)$ exists.
To see that this limit exists, just fix some $x>0$ and use that
$$g(t)=\frac{h(\sqrt{x^2+t^2})}{f(x)}$$
which implies
$$\lim\limits_{t\to0^+} g(t)=\lim\limits_{t\to0^+} \frac{h(\sqrt{x^2+t^2})}{f(x)}=\frac{h(x)}{f(x)}$$
by continuity of the function $h$. This shows that the limit used in the previous argument indeed exists. And it also gives a different argument that $h(x)/f(x)$ is constant.
A: This should start you off:
$$f(x)g(0)=h(\sqrt{x^2+0^2}) = h(x)$$
Thus 
$\frac{h(x)}{f(x)}=g(0)$
Is a constant
However $g$ is not defined at 0 so instead we consider the following function and limit (assuming the functions are continuous)
$$\quad\alpha(a;x)=2xa+a^2$$
$$\Leftrightarrow x+a=\sqrt{x^2+\alpha(a;x)^2}$$
So then we must have
$$\frac{h(x)}{f(x)}=\lim_{a\downarrow0}\frac{h(x+a)}{f(x)}=\lim_{a\downarrow0}\frac{f(x)g(\alpha(a;x))}{f(x)}$$
I'm sure you can figure out the rest. There is an argument to be made about the existence of this limit (it must converge or be unbounded and if it's unbounded then $g\to\infty$ as $x\to0$.
You can do the same argument for the middle case and then combine the two for the first case.
