MathSaiyan wrote: Find all functions $f\colon \mathbb{R}^+ \to \mathbb{R}^+$ such that \[ f(xf(x)+y^2) = x^2+yf(y) \]for any positive reals $x,y$. Let $P(x,y)$ be the assertion $f(xf(x)+y^2)=x^2+yf(y)$ Taking $f()$ both sides of $P(x,y)$, we get $f(f(xf(x)+y^2))=f(x^2+yf(y))$ But $P(y,x)$ $\implies$ $f(x^2+yf(y))=y^2+xf(x)$ And so $f(f(xf(x)+y^2))=xf(x)+y^2$ And so $f(f(x))=x$ $\forall x>a$ where $a=\inf_{x\in\mathbb R^+}(xf(x))$ Let then $x>a$. Comparing $P(x,y)$ with $P(f(x),y)$, we get $f(x)^2=x^2$ And so $f(x)=x$ $\forall x>a$ Choosing then $y$ such that both $y>a$ (so that $f(y)=y$) and $y^2>a$ (so that $f(xf(x)+y^2)=xf(x)+y^2$), $P(x,y)$ becomes $xf(x)+y^2=x^2+y^2$ And so $\boxed{f(x)=x\quad\forall x>0}$, which indeed fits

pco wrote: Choosing then $y$ such that both $y>a$ (so that $f(y)=y$) and $y^2>a$ (so that $f(xf(x)+y^2)=xf(x)+y^2$), $P(x,y)$ becomes $xf(x)+y^2=x^2+y^2$ It is even simpler than this. $xf(x) + y^2 > a$ for any $x$ and $y$ by definition of $a$

I think my solution is easier but more difficult to come up with. I first define the function $g(x)=xf(x)$. We observe that this new function is injective because $f(g(x)+y^2)=x^2+g(y)$, and if we fix $y$ and set $x\neq x'$ such that $g(x)=g(x')$, $x^2+g(y)=f(g(x)+y^2)=f(g(x')+y^2)=(x')^2+g(y)\implies x=x'$, contradiction. Moreover, we can rewrite the functional equation in terms of $g$, obtaining $g(g(x)+y^2)=(x^2+g(y))(y^2+g(x))=g(g(y)+x^2)$ because of the symmetric form of the RHS expression, due to $g$'s injectivity, $g(g(x)+y^2)=g(g(y)+x^2)\implies g(x)+y^2=g(y)+x^2\implies x^2-g(x)=k=y^2-g(y)$ for some real $k$, so $g(x)=x^2-k\implies f(x)=x-\frac{k}{x}$. But setting $x=y$, $f(xf(x)+x^2)=x^2+xf(x)$, so there exist positive reals $x$ such that $f(x)=x=x-\frac kx\implies k=0\implies f(x)=x$ is the only solution.