Functional equation with sqrt solutions Let $f:(0,\infty)\to(0,\infty)$ so that for all $x,y\in(0,\infty)$ we have
$$f\left(\frac {x} {f(y)}\right)=\frac {x} {f(x\sqrt y)}.$$
Find function $f$.
 A: Let $P(x,y)$ be the assertion $f\left(\frac{x}{f(y)}\right)f(x\sqrt{y})=x$ which is equivalent to $f\left(\frac{x}{f(y)}\right)=\frac{x}{f(x\sqrt{y})}$.
Then:
$$
P(xf(1),1)\ :\ f(x)f(xf(1))=xf(1)\implies f(f(1))=1\ \text{ and }\ f(f(1)^2)=f(1)^2 \\
P(x,f(1))\ :\ f(x)f\left(x\sqrt{f(1)}\right)=x
$$
Thus we obtain on the one hand: $f(x)f\left(x\sqrt{f(1)}\right)f(xf(1))=xf(xf(1))$ but one the other hand we have $f\left(x\sqrt{f(1)}\right)f(xf(1))=x\sqrt{f(1)}$ and thus $f(x)f\left(x\sqrt{f(1)}\right)f(xf(1))=f(x)x\sqrt{f(1)}$. Be combining these two equations, we get:
$$
xf(xf(1))=f(x)x\sqrt{f(1)}\implies f(xf(1))=f(x)\sqrt{f(1)}
$$
By plugging in $x=1$ in this equation, we obtain $f(1)^{\frac{3}{2}}=1$ and thus $f(1)=1$. Finally we obtain:
$$
P(x,1)\ :\ f(x)^2=x\implies f(x)=\sqrt{x}
$$
for all $x$ in $(0;\infty)$, which is indeed a solution.
A: $f(x) = \sqrt{x}$ satisfies the property.
Let $x,y \in (0,\infty)$.
Then
$$f\left(\frac{x}{f(y)}\right) = \sqrt{\frac{x}{\sqrt{y}}} = \frac{\sqrt{x}}{\sqrt[4]{y}} = \frac{x}{\sqrt{x}\sqrt[4]{y}} = \frac{x}{\sqrt{x\sqrt{y}}} = \frac{x}{f(x\sqrt{y})}$$
