Constant functions

forall $x,y,z>0$, $h(\sqrt{x^2+y^2})f(z) = f(x) g(y) f(z) = f(x) h(\sqrt{y^2+z^2})$,
thus $\frac {h(\sqrt{x^2+y^2})}{h(\sqrt{z^2+y^2})} = \frac{f(x)}{f(z)} $.

Therefore, forall $x,y,z,t > 0$ : $\frac {h(\sqrt{x^2+z^2})}{h(\sqrt{y^2+z^2})} = \frac {f(x)}{f(y)} = \frac {h(\sqrt{x^2+z^2+t^2})}{h(\sqrt{y^2+z^2+t^2})} = \frac {f(\sqrt{x^2+z^2})}{f(\sqrt{y^2+z^2})}$, thus $\frac {h(\sqrt{x^2+z^2})}{f(\sqrt{x^2+z^2})} = \frac{h(\sqrt{y^2+z^2})}{f(\sqrt{y^2+z^2})} $, which proves that $h/f$ is a constant function.

We may assume $f(1)=g(1)=1$. It follows that $f(x)=h\bigl(\sqrt{x^2+1}\bigr)=g(x)$ for all $x>0$. Put $$H(t):=h\bigl(\sqrt{t}\bigr)\qquad(t>0)\ ,$$ then $$f(x)\ f(y)=H(x^2+y^2)\qquad(x>0, \ y>0)\ .$$ Taking logarithms we obtain $\log f(x)+\log f(y)=\log H(x^2+y^2)$ or $$\log H(x^2+1)+\log H(y^2+1)=\log H(x^2+y^2)\ .\qquad(1)$$ We now write $x^2:=1+u$, $\ y^2:= 1+v$ with $u$ and $v$ near $0$ and introduce the new function $\phi(t):=\log H(2+t)$. Then $(1)$ becomes the familiar functional equation $$\phi(u)+\phi(v)=\phi(u+v)\ .\qquad(2)$$ If we insist that $f$, $g$, $h$ are continuous then the only solutions to $(2)$ are the functions $\phi(t)=C\,t$, and going all the way backwards the claim about $f$, $g$, $h$ follows.