pomodor_ap wrote:

Let $f : \mathbb{R}^+ \to \mathbb{R}^+$ be a function such that
$f(x)f(x^2 + y f(y)) = f(x)f(y^2) + x^3$ for all $x, y \in \mathbb{R}^+$ . Determine all such functions $f$ .

Very nice.

Claim 1: $f(x)\ge x$
Suppose $f(y)<y$ for some $y$ . Setting $x=\sqrt{y^2-yf(y)}$ so that $f(x)f(x^2+yf(y))=f(x)f\left(y^2\right)$ , we get:
$\left(\sqrt{y^2-yf(y)}\right)^3=0\Rightarrow f(y)=y,$ contradiction.

Now taking $x\mapsto\sqrt x$ , $y\mapsto\sqrt y$ in the original equation, we get:
$f\left(x+\sqrt yf\left(\sqrt y\right)\right)=\frac{x^{3/2}}{f\left(\sqrt x\right)}+f(y),$ so defining $g(x)=\sqrt xf\left(\sqrt x\right)$ and $h(x)=\frac{x^{3/2}}{f\left(\sqrt x\right)}$ , we have:
$f(x+g(y))=h(x)+f(y).$ Note that $f(1)=g(1)$ , and by Claim 1, $h(x)\le x$ .

Claim 2: $g(x)=f(x)$
First, we have:
$f(g(x)+g(y))=h(g(x))+f(y)$ and switching $x,y$ , we get that $h(g(x))=f(x)+c$ for some constant $c\in\mathbb R$ .
Since $h(x)\le x$ this gives $g(x)\ge f(x)+c$ .

So this equation becomes $f(g(x)+g(y))=f(x)+f(y)+c$ , and since $f(x)\ge x$ we have:
$g(x)+g(y)\le f(x)+f(y)+c.$ Setting $y=1$ we get $g(x)\le f(x)+c$ .

Because of these two, $g(x)=f(x)+c$ for all $x\in\mathbb R^+$ . Since $g(1)=f(1)$ again, we have $c=0$ , hence proven.

Claim 3: $h(x)=x$
Trivial:
$x+f(y)\le f(x+f(y))=f(x+g(y))=h(x)+f(y)\le x+f(y)$ with equality only when $h(x)=x$ .

Then $\frac{x^{3/2}}{f\left(\sqrt x\right)}=x$ , giving the unique solution $\boxed{f(x)=x}$ which fits.