Functions $f(x,y)$ such that $\left[f\left(x^{\frac{1}{t}},y^{\frac{1}{t}} \right)\right]^t$ does not depend on $t$? Function $f(x,y)$ is defined for $x>0$ and $y>0$. It satisfies the following property: $\left[f\left(x^{\frac{1}{t}},y^{\frac{1}{t}} \right)\right]^t$ does not depend on $t$ $\forall t\ne0,x>0,y>0$. Obviously, $f(x,y)=x^{a_x}y^{a_y}$ satisfies this property. Are there any other functions like that? 
 A: Assume
$$f(x^r,y^r)=f(x,y)^r$$
for all positive real $x,y,r$. By letting
$$g(x,y)=\log f(e^x,e^y),$$
we see that
$$e^{g(\log x,\log y)}=f(x,y),$$
so our condition becomes
$$e^{g(\log x^r,\log y^r)}=e^{rg(\log x,\log y)},$$
so $g$ is a function from $\mathbb R\times\mathbb R\to\mathbb R$ satisfying
$$g(ra,rb)=rg(a,b)$$
for all real $r,a,b$. 
We have for any $k$ that
$$g(r,rk)=rg(1,k),$$
and so letting $h(k)=g(1,k)$ we can represent
$$g(x,y)=\begin{cases}xh(y/x)&\text{if }x\neq 0 \\ yg(0,1)&\text{otherwise}.\end{cases}$$
However, these are all the conditions we need -- for any function $h:\mathbb R\to\mathbb R$ and constant $c=g(0,1)$ we can define such a $g$ and then an $f$, generating many different functions $f$ that work. Most of them will not be very easily describable.
One nice function, for example, comes from
$$g(x,y)=\sqrt{x^2+y^2}.$$
A: So in particular,
$$\left[f\left(x^{\frac{1}{t}},y^{\frac{1}{t}} \right)\right]^t=f(x,y).$$
If $x=y=1$, we obtain that $f(1,1)^t$ does not depend on $t$. We conclude
$$ f(1,1)=1.$$
If $x\ne 1$, plug in $t:=\log_2x$ to arrive at
$$ f(x,y)=\left[f\left(2,y^{\frac{1}{\log_2x}} \right)\right]^{\log_2x}=x^{\log_2f\left(2,y^{\frac{1}{\log_2x}} \right)}.$$
If $x=1$ and $y\ne 1$, plug in $t:=\log_2y$ to arrive at
$$ f(1,y)=\left[f\left(1,2 \right)\right]^{\log_2y}=y^{\log_2 f(1,2)}$$
In summary, $f$ has the desired property iff it is of the form
$$f(x,y)=\begin{cases}1&x=y=1\\
x^{g\left(y^{\frac1{\log_2x}}\right)}&x\ne1\\
y^\alpha&x=1\ne y\end{cases} $$
