mathwizard888 wrote:

Find all functions $f:(0,\infty)\rightarrow (0,\infty)$ such that for any $x,y\in (0,\infty)$ , $xf(x^2)f(f(y)) + f(yf(x)) = f(xy) \left(f(f(x^2)) + f(f(y^2))\right).$

Answer: The only functions which work are $f(x)=\tfrac{1}{x}$ for all $x>0$ .

Let $P(x, y)$ denote the given functional equation: $xf(x^2)f(f(y))+f(yf(x))=f(xy)\cdot \left(f(f(x^2))+f(f(y^2))\right),$ then it is easy to see that $f(x) \equiv \tfrac{1}{x}$ is indeed a solution. We now show that this is the only one.

Note that $P(1, 1) \implies f(1)=1$ and $P(1, y) \implies f(f(y))=f(y)f(f(y^2)),$ for all $y>0$ . Also, $P(x, 1) \implies xf(x^2)=f(x),$ for all $x>0$ . Hence, we can rewrite $P$ as $f(x)f(f(y))+f(yf(x))=f(xy) \cdot \left(\frac{f(f(x))}{f(x)}+f(f(y^2))\right).$
Claim: $f$ is injective.

(Proof) Suppose $f(a)=f(b)$ for some $a \ne b$ . Together, $P(a, y)$ and $P(b, y)$ yield that $f(ay)=f(by)$ for all $y>0$ ; hence, for $t=\tfrac{1}{a}$ we have $f(t)=f(1)=1$ . However, this yields $f(t^2)=f(t \cdot t)=f(t)=1,$ so $f(t)=tf(t^2)=tf(t) \implies t=1,$ hence $\boxed{a=b}$ . $\blacksquare$

Finally, we observe that $f\left(f(x)^2\right)=\frac{f(f(x))}{f(x)}=f(f(x^2))=f\left(\frac{f(x)}{x}\right) \implies f(x)^2=\frac{f(x)}{x} \implies f(x) \equiv \frac{1}{x},$ due to injectivity, hence we conclude that the initial assertion is valid. $\blacksquare$