Find all functions $f:\mathbb{R}_{\geq{0}} \times \mathbb{R}_{>{0}} \to \mathbb{R}_{\geq{0}}$ such that for all $x \in \mathbb{R}_{\geq{0}}$ and all $y, z \in \mathbb{R}_{>{0}},$ $$ f(f(x,y),z)=f(x,yz)$$ and $$\forall x\in\mathbb{R}_{\geq{0}}: f(x,1)=x$$ where the function $f$ is differentiable over its entire domain (at least once).
I found a solution assuming that the function is analytical. Then by unfolding into Taylor's series and comparing the coefficient, I got $$f(x,y) = y^{\alpha}x$$ where $\alpha$ is any real number.
I don't know how to prove that this solution is unique and how to circumvent the proof without the additional requirement of analytical function.
Edit
I already know why I had a problem with the ambiguity of the function. I forgot a very important condition.
Function $f$ is homogenous with respect to $x$: $$f(\lambda{}x,y)=\lambda{}f(x,y)$$ Then there is only one unique solution: $f(x,y)=xy^\alpha$.
Thank you all for your advice, primarily @Max for his guidance to Transaltion function as general solution without this homogenous condition.
Let $F(x,y)=\ln f(e^x, e^y)$ . Then
$$\begin{aligned} F(F(x,y), z)&=\ln f(e^{\ln f(e^x, e^y)}, e^z)= \\&\ln f( f(e^x, e^y), e^z)=\ln f(e^x, e^{y+z})\\&=F(x, y+z). \end{aligned}$$
This is the translation equation (see https://en.wikipedia.org/wiki/Abel_equation#Equivalence). Pick arbitrary continuous and strictly monotone $h$ with range all of $\mathbb{R}$ (and differentiable at least once with positive derivative, to get corresponding smoothness of $f$). Then $F(x, y)=h(h^{-1}(x)+y)$ solves it. Then $f(x,y)=e^{F(\ln x, \ln y)}=e^{ h(h^{-1}(\ln x)+\ln y)}$
Your solution is obtained when $h(u)=\alpha u$, i.e. $f(x,y)=e^{\alpha (\frac{1}{\alpha} \ln x+\ln y)}=x y^{\alpha}$.
The solution $f(x,y)=x^y$ is almost obtained from $h(u)=e^{u}$: if one ignores the fact that $\ln \ln x$ is not defined for $x\leq 1$ we have $f(x,y)=e^{e^{\ln \ln x+ \ln y}}=e^{y\ln x}=x^y$.
There is a discussion of translation equation in Section 6.1 of Aczel's book on functional equations. It shows that (under mild continuity conditions) if $F(x,\cdot)$ is non-constant for every $x$ then the solution is in fact of the form above. This suggests that $f(x,y)=x^y$ is in some sense exceptional.