Solving a functional relation $f\left( x \cdot f(y)\right)=x^2 \cdot y^a$ I have this functional relation - $$f\left( x \cdot f(y)\right)=x^2 \cdot y^a$$ which I am trying to solve. 
I put $x=1$, then I put $f(y)=\dfrac{1}{x}$. I also tried out $y=f^{-1}(1)$, but it doesn't seem to work out. Please help me out. Thank you.
 A: For $x>0$:
$$f(f(x))=x^a$$
$$f(\sqrt x f(1))=x$$
so
$$f(x)=f(f(\sqrt xf(1)))=\sqrt{x^a}f(1)^a$$
This would determine $f$, chosen $f(1)$ and $a$.
Now, let's see if we really can choose:
$$x^2y^a=f(xf(y))=f(x\sqrt{y^a}f(1)^a)=x^{a/2}y^{a^2/4}f(1)^{3a/2}$$
For $y=1$:
$$x^4=x^af(1)^{3a}\quad\forall x>0$$
hence $x^{4-a}$ is constant, so it must be $a=4$, $f(1)=1$. So finally
$$f(x)=x^2$$
A: OK, I make my comment more precise. Put $y=1$, you get $f(xf(1))=x^2$. Hence $f(1)\not =0$, (if $f(1)=0$, then $x^2=f(0)$ for all $x$, a contradiction). If I put $z=xf(1)$, I get $f(z)=\frac{z^2}{(f(1))^2}$ for all $z$. I put $z=1$, it gives $f(1)^3=1$, hence $f(1)=1$. We have proven that if a solution exists, then it is $f(x)=x^2$. Hence: if $a\not =4$, there is no solutions. If $a=4$, there is only one solution, $f(x)=x^2$.
A: If $f$ is invertible you can do:
\begin{equation}
x \cdot f(y) = f^{-1}(x^{2} \cdot y^{a})
\end{equation}
if $x = 0$:
\begin{equation}
0 \cdot f(y) = f^{-1}(0 \cdot y^{a}) => f^{-1}(0) = 0 => f(0) = 0 
\end{equation}
than if $x \ne 0$:
\begin{equation}
f(y) = \frac{f^{-1}(x^{2} \cdot y^{a})}{x}
\end{equation}
than you can put $x = 1$
\begin{equation}
f(y) = f^{-1}(y^{a}) => f(f(y)) = y^{a}
\end{equation}
$=> f(y)$ can be $y^{\sqrt a}$ but this solution don't solve the general case.
I think that solve the problem for any $x$ you must have another condition to solve.
A: I think this approach may be of interest, too:
Assume $ f : \mathbb R \to \mathbb R $ satisfies
$$ f \big( x f ( y ) \big) = x ^ 2 y ^ a \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $. Put $ x = y = 1 $ in \eqref{0} to get $ f \big( f ( 1 ) \big) = 1 $. Now Let $ y = f ( 1 ) $ in \eqref{0} and you'll have
$$ f ( x ) = x ^ 2 f ( 1 ) ^ a \text . \tag 1 \label 1 $$
Substitute $ f ( x ) $ for $ x $ in \eqref{0} to get
$$ f \big( f ( x ) f ( y ) \big) = f ( x ) ^ 2 y ^ a \text . \tag 2 \label 2 $$
By symmetry, \eqref{2} leads to
$$ f ( x ) ^ 2 y ^ a = x ^ a f ( y ) ^ 2 \tag 3 \label 3 $$
for all $ x , y \in \mathbb R $. Putting $ y = 1 $ in \eqref{3} we'll have
$$ f ( x ) ^ 2 = x ^ a f ( 1 ) ^ 2 \tag 4 \label 4 $$
for all $ x \in \mathbb R $. Now, we can compare \eqref{1} and \eqref{4}to get
$$ x ^ a f ( 1 ) ^ 2 = x ^ 4 f ( 1 ) ^ { 2 a } \tag 5 \label 5 $$
for all $ x \in \mathbb R $. In particular, this shows that
$$ f ( 1 ) ^ { 2 a } = f ( 1 ) ^ 2 \text , \tag 6 \label 6 $$
and \eqref{5} simplifies to
$$ f ( 1 ) ^ 2 \left( x ^ a - x ^ 4 \right) = 0 \text . \tag 7 \label 7 $$
Assuming $ a \ne 4 $, we can put $ x = 2 $ in \eqref{7} and conclude that $ f ( 1 ) = 0 $. But this is impossible, since then we can let $ y = 1 $ in \eqref{0} to get $ f ( 0 ) = x ^ 2 $ for all $ x \in \mathbb R $, which leads to a contradiction. Therefore we can't have $ a \ne 4 $, and we must have $ a = 4 $. Using \eqref{6} and the fact that $ f ( 1 ) $ cannot be equal to $ 0 $, this shows that $ f ( 1 ) ^ 2 = 1 $, and finally by \eqref{1}, we get $ f ( x ) = x ^ 2 $ for all $ x \in \mathbb R $. It's straightforward to check that this satisfies \eqref{0}, and is indeed a solution.
