<s>[INST] Find all functions such that $f(x^2+y^2f(x))=xf(y)^2-f(x)^2$ I am dealing with the test of the OBM (Brasilian Math Olympiad), University level, 2016, phase 2.
I hope someone can help me to discuss this test. Thanks for any help.
The question 2 says:

Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ such that 
$f(x^2+y^2f(x))=xf(y)^2-f(x)^2$
for all $x,y\in\mathbb{R}$.

My attempt:
Note that $f(0)\in\{0,-1\}$. In fact, by taking $x=y=0$, we have $f(0)=-f(0)^2$.
Case 1 $f(0)=0$
By taking $y=0$, we have 
$f(x^2)=-f(x)^2\forall x\in\mathbb{R}$
Particularly, $f(1)=-f(1)^2$, so $f(1)\in\{0,-1\}$.
(a) f(1)=0
By taking $x=1$, we have $f(1)=f(y)^2\forall y\in\mathbb{R}$.
So, $f\equiv 0$. Is trivial that it respects the statement.
(b) f(1)=-1
By taking $x=1$, we have $f(1-y^2)=f(y)^2-1=-f(y^2)-1\forall y\in\mathbb{R}$. So, to $t\leq 0$, we have $f(1-x)=-f(x)-1$.
By taking $y=1$, we have $f(x^2+f(x))=x-f(x)^2=x+f(x^2) \forall x\in\mathbb{R}$. 
I could not finish this subcase
Case 2 $f(0)=-1$
By taking $x=0$,
$f(-y^2)=-1\forall y\in\mathbb{R}$
So, $f(t)=-1\forall t\leq0$.
By taking $y=0$,
$f(x^2)=-x-f(x)^2 \forall x\in\mathbb{R}$
So, $f(t)=-\sqrt{t}-1\forall t\geq0$.
But this function does not is correct. For instance, to $x=y=1$, $f(x^2+y^2f(x))=f(1+1(-2))=f(-1)=-1$, but $xf(y)^2-f(x)^2=1(-2)^2-(-2)=6\not=-1$.
 [/INST] Partial progress, but not a complete answer, I'm afraid.
$$f(x^2+y^2f(x)) = xf(y)^2-f(x)^2$$
$f$ has a root
Let $y=x$; then $f(x^2(1+f(x)) = (x-1)f(x)^2$. In particular, letting $x=1$ we obtain $f(1+f(1)) = 0$, so $f$ does have a root.
$f$ is $0$ or has exactly the root $0$
Suppose $f(x) = 0$. Then $f(x^2) = x f(y)^2$ for all $y$, and so either $x = 0$ or $f(y)^2$ is constant as $y$ varies.
Suppose $f(x) = 0$ but $x \not = 0$. Then $f(y)^2$ is constant as $y$ varies; but substituting $y = x$ we obtain that $f(y)^2 = 0$ and hence $f$ is the constant $0$.
So the only possible nonzero case is that $f$ has exactly one root, and it is the root $x = 0$.
$f$ is very nearly symmetric
Substitute $y \to -y$ to obtain the following:
$$x f(y)^2-f(x)^2 = f(x^2+y^2f(x)) = x f(-y)^2-f(x)^2$$
from which $$x f(y)^2 = x f(-y)^2$$ for all $x$ and $y$; in particular, $$f(y) = \pm f(-y)$$ for all $y$.
$f$ is odd or $0$
Suppose $f(x) = f(-x)$. Then $$x f(y) - f(x)^2 = f(x^2 + y^2 f(x)) = -x f(y) - f(-x)^2 = -x f(y) - f(x)^2$$
and so $-x f(y) = x f(y)$ for all $y$; so (since wlog $f$ is not the constant zero function) $-x = x$ and hence $x=0$.
So if $f(x) = f(-x)$ then $x = 0$; hence $f(-x) = -f(x)$ for all $x$.
$f$ is sign-reversing or $0$
Note also that since $f(x^2) = -f(x)^2$ (by letting $y=0$), for every $x > 0$ we have $f(x) < 0$.
$f(n) = -n$ or $f=0$
Substituting $x=-1$ gives $f(1+y^2) = -f(y)^2-1$
and in particular $$f(x^2+1) = f(x^2)-1$$
Therefore $f(x+1) = f(x)-1$ whenever $x>0$.
This fixes the value of $f$ on the natural numbers: we have $f(n) = -n$.

We already know that the root occurs at $x=1+f(1)$, so $f(1) = -1$ (as you noted). Moreover, by letting $x=y$ and supposing $f(x)=-1$, we get $f(0) = x-1$ at any such $x$, and so $x=1$ is the only time $f$ hits $-1$.
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