Functional equation $(1-z)f(x)=f(\frac{1-z}{z}f(xz))$

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May you could help me with the following functional equation: $(1-z)f(x)=f(\frac{1-z}{z}f(xz))$. I want to find all function $f:[0,\infty)\rightarrow[0,\infty)$ for $x>0,0<z<1$

Its an exercise from the American Monthly Problems from July 2012. My idea was to take the derivative (with repsect to z), but it did not help.

EDIT1:As stated in the comment, my first mistake was to assume differentiability. Now I consider two cases, $x=0$ and $z=1$.

$z=1$ => $0=f(0)$ and $x=0$ => $(1-z)f(0)=f(\frac{1-z}{z}f(0))$ => $0=f(0)$

I do not see where I got some new informations, now I only now that $0=f(0)$

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Derivatives would not help you anyway, since you don't have any reason to believe that $f(x)$ is differentiable. The key to such problems is always to use special values for $x,z$ and draw conclusions from those simpler equations.