Solving the functional equation $f\left(y^2f(x)+x^2f(y)\right)=xy\big(f(x)+f(y)\big)$

Question

Problem: find all continuous functions $f:[0,+\infty)\to [0,+\infty)$ such that $$f\left(y^2f(x)+x^2f(y)\right)=xy\big(f(x)+f(y)\big),\;\forall x,y\in [0,+\infty)\text.$$

Answer 1

Problem:find all continuous functions $f:[0,+\infty)\to [0,+\infty)$ such that $$f(y^2f(x)+x^2f(y))=xy(f(x)+f(y)),\;\forall x,y\in [0,+\infty)$$
Solution:

We see that $ f(x)=0 $ is one solution.

We now proceed to find other solutions that are not identically zero. For these solutions we can find points $ a $ for which $ f(a) \neq 0 $.

Also note that $ f(0)=0 $ :

$f(0)=f(0f(0)+0f(0))=0(f(0)+f(0))=0 $.

$ f(x) $ must be surjective:

Choose $ a \neq 0 $ for which $ f(a) \neq 0 $. Now : $f(y^2f(a)+a^2f(y))=ay(f(a)+f(y)) \ge yaf(a) ,\;\forall y\in [0,+\infty)$ . The right part of this inequality can be made arbitrarily large. So values of $ f(x) $ range from $ 0 $ to $ \infty $. Because of the continuity requirement all values in between must be reached somewhere $ \implies f(x) $ is surjective.
Also the argument of $ f() $ on the left of the inequality above: $ y^2f(a)+a^2f(y) $ goes to $ 0 $ when $ y $ goes to $ 0 $ , and it goes to $ \infty $ when $ y $ goes to $ \infty $.

Because of the surjectivity, continuity and the fact that $ f(0)=0 $ the function: $ \lambda(x) = 2x^2f(x) $ is also continuous, surjective and ranges from $ 0 $ to $ \infty $.
In other words: every $ a \in \mathbb{R_{\ge 0}} $ can be written as $ 2b^2f(b) $ for some $ b \ge 0 $ .

Therefore we can change the coordinates of $ f() $ from $ x $ to $ \lambda(x) = 2x^2f(x) $ without loss of generality.

$f(\lambda)=f(2x^2f(x)) =f(x^2f(x)+x^2f(x))=x^2(f(x)+f(x))=2x^2f(x)=\lambda\implies $

The only continuous solutions are : $ f(x)=0 $ and $ f(x)=x $.