Suppose $f:\Bbb R\to \Bbb R$ satisfies
$$f\big(f(x)-x+y^2\big)=yf(y)\tag{1}$$
for all $x,y\in \Bbb R$. Plugging in $x,y=1$ in $(1)$, we get
$$f\big(f(1)\big)=f(1).$$
Hence, substituting $x=f(1)$ in $(1)$, we obtain
$$f(y^2)=yf(y).\tag{2}$$
This shows that $f(0)=0$ and for $y\ne 0$,
$$-yf(-y)=f\big((-y)^2\big)=f(y^2)=yf(y),$$
so $f(-y)=-f(y)$. Hence $f$ is an odd function.
From $(1)$ and $(2)$, we get
$$f\big(f(x)-x+y^2\big)=f(y^2)$$
for all $x,y\in\mathbb{R}$. Substitute $-x$ for $x$ in the last equation and using the fact that $f$ is odd, we get
$$f\big(f(x)-x-y^2\big)=f(-y^2)$$
for all $x,y\in\mathbb{R}$. Hence,
$$f\big(f(x)-x+y)=f(y)\tag{3}$$
for all $x,y\in\mathbb{R}$.
Define $P$ to be the additive subgroup of $\mathbb{R}$ generated by $\big\{f(x)-x\big|x\in\Bbb R\big\}$. Then, we see that for any $z\in\mathbb{R}$ and $p\in P$, we have
$$f(z+p)=f(z).\tag{4}$$
Consequently $f$ is an odd periodic function which is invariant under translation by elements of $P$.
We claim that $P=f^{-1}(0)$. First if $p\in P$, then $f(p)=f(p+0)=f(0)=0$, so $p\in f^{-1}(0)$. Conversely, suppose that $p\in f^{-1}(0)$, then with $x=p$ in $(3)$, we have
$$f(-p+y)=f\big(f(p)-p+y\big)=f(y)$$
so that $p\in P$.
For each $p\in P$, we see that
$$f\left(\left(\frac{1+p}{2}\right)^2\right)=f\left(\left(\frac{1-p}{2}\right)^2+p\right)=f\left(\left(\frac{1-p}{2}\right)^2\right)$$
From $(2)$, we have
$$\frac{1+p}{2}f\left(\frac{1+p}{2}\right)=\frac{1-p}{2}f\left(\frac{1-p}{2}\right).$$
However
$$f\left(\frac{1+p}{2}\right)=f\left(\frac{1-p}{2}+p\right)=f\left(\frac{1-p}{2}\right).$$
This means
$$f\left(\frac{1+p}{2}\right)=0$$
for all $p\in P$ such that $p\ne 0$. Hence, for $p\in P\setminus\{0\}$, $\frac{1+p}{2}\in P$, so $1+p \in P$, making $1\in P$. This shows that either $P=\{0\}$ or $\mathbb{Z}\left[\frac12\right]\subseteq P$.
If $P=\{0\}$, then $f(x)-x=0$ for all $x\in \Bbb R$. Therefore, $f(x)=x$ for every $x\in \Bbb R$. This yields one solution of $(1)$. From now on $P\neq \{0\}$. Thus $\mathbb{Z}\left[\frac12\right]\subseteq P$. We want to show that $P=\Bbb R$, so that $f(x)=0$ for every $x\in \Bbb R$, and this is another solution of $(1)$.
For an arbitrary $y\in\mathbb{R}$, we see that
$$f\left(\left(y+\frac{1}{2}\right)^2\right)=\left(y+\frac{1}{2}\right)f\left(y+\frac{1}{2}\right)=\left(y+\frac{1}{2}\right)f(y),$$
since $1/2\in P$. That is
$$f\left(\left(y+\frac12\right)^2\right)=\frac{1}{2}f(y)+yf(y)=\frac12f(y)+f(y^2).$$
Consequently
\begin{align}f\left(\left(y+\frac12\right)^2\right)-\left(y+\frac12\right)^2&=\frac{1}{2}f(y)+f(y^2)-\left(y+\frac12\right)^2\\&=\left(\frac{1}{2}f(y)-y\right)+\left(f(y^2)-y^2\right)+\frac14.\end{align}
Because $f\left(\left(y+\frac12\right)^2\right)-\left(y+\frac12\right)^2$, $f(y^2)-y^2$, and $\frac14$ are all elements of $P$, we conclude that
$$\frac{1}{2}f(y)-y\in P.$$
Therefore,
$$\big(f(y)-y\big)-y=2\left(\frac{1}{2}f(y)-y\right)\in P.$$
As $f(y)-y\in P$, we get $y\in P$.
Best Answer
Suppose that $f:\mathbb{R}\to\mathbb{R}_{\geq 0}$ is a function such that $$f(x^2+y^2)=f(x^2-y^2)+f(2xy)\,,\tag{*}$$ for all $x,y\in\mathbb{R}$. For each $x,y\in\mathbb{R}$, let $P(x,y)$ denote the statement (*).
From $P(0,0)$, we get $f(0)=0$. Thus, $P(0,y)$ yields $$f(+y^2)=f(-y^2)$$ for all $y\in\mathbb{R}$. Therefore, $f$ is an even function. Let $F:\mathbb{R}_{\geq 0}\to\mathbb{R}$ be the function defined by $$F(t):=f(\sqrt{t})$$ for all $t\geq 0$.
Now, since $f$ is an even function, $$\begin{align}F\big((x^2+y^2)^2\big)&=f(x^2+y^2)=f(x^2-y^2)+f(2xy)\\&=f\big(|x^2-y^2|\big)+f\big(2|xy|\big)\\&=F\big((x^2-y^2)^2\big)+F\big((2xy)^2\big)\end{align}$$ for all $x,y\in\mathbb{R}$. This shows that $$F(u+v)=F(u)+F(v)\tag{#}$$ for all $u,v\geq 0$. (Note that, for $u,v\geq 0$, there are $x,y\in\mathbb{R}$ such that $u=(x^2-y^2)^2$ and $v=(2xy)^2$.) Thus, $F$ satisfies Cauchy's functional equation and $F$ is a nonnegative function. Therefore, $F$ is a nondecreasing function. Ergo, there exists $k\geq 0$ such that $$F(t)=kt$$ for all $t\in\mathbb{R}_{\geq 0}$.
Since $f(x)=f\big(|x|\big)=F(x^2)$ for all $x\in\mathbb{R}$, we conclude that $$f(x)=kx^2$$ for all $x\in\mathbb{R}$ (where $k\geq 0$ is a constant). It is easy to see that all functions $f$ of this form obey (*).
Remark. For a function $F:\mathbb{R}_{\geq 0}\to\mathbb{R}$ such that $F$ satisfies Cauchy's functional equation (#), we can conclude that there exists a constant $k$ such that $F(t)=kt$ for all $t\geq 0$ if at least one of the following properties are known to be true:
Without these properties, there are noncontinuous examples of $F$ (but such examples rely on the Axiom of Choice).