proof verification - Functional equation: $f(x^2+f(y))=y+f^2(x)$.

Functional Equation:

Solve for a given function $f:R\rightarrow R$, $$f(x^2+f(y))=y+f^2(x)\tag{$\ast$}$$
Note: $f^2(x)=f(x)\cdot f(x)$.

Not sure if the proof is correct (I have made quite a few fake proofs):

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My Attempt:

Fix $x$ to show $f$ is surjective, and assume $f(y_1)=f(y_2)$. Then, sub $y=y_1$ and $y=y_2$ to show $f$ is injective. Therefore, $f$ is bijective.

One may find a $k$ such that $f(k)=0$. Sub into $(*)$, giving $$f(k^2+f(y))=y.$$ Now, one may find $m$ such that $f(m)=-k^2$. Sub it in to obtain $f(0)=m$. Therefore, we have the following simultaneous equation: $$\begin{align}f(f(0))&=-k^2 \\ \text{and }\qquad f(k)&=0.\end{align}$$ Set $x=y=0$, giving $f(f(0))=f^2(0)$. So we have $f^2(0)=-k^2$. Since the LHS$\,\ge0$ and RHS$\,\le0$, then $f^2(0)=-k^2=0$ and thus, $f(0)=0$.

Now set $x=0$ to obtain that $f(f(y))=y$. In other words, $f$ is an involution.

Now set $y=0$ to obtain $f(x^2)=f^2(x)$. Thus, $f^2(-x)=f(x^2)=f^2(x)$. Since $f$ is injective, then $f(x)\neq f(-x)$. Therefore, $f(x)=-f(-x)$. In other words, $f$ is odd.

Set $y=f(x^2)$ in $(*)$ to obtain that $f(2x^2)=2f(x^2)$. Set $x=1$ and $y=0$ in $(*)$ to obtain $f(1)=1$. Set $c=x^2$ to obtain $f(2c)=2f(c)$. Therefore, $$\forall x>0, \ f(x)=x\text{ is the only solution.}$$ Since $f$ is an odd function, I claim that $f(x)=x$ is certainly the only solution.

Edit: new last step
Sub $x=f(z)$ and $y=0$ in $(*)$ gives $$f(z^2)=z^2$$ Since $f$ is an odd function, $f(x)=x$ is the only solution.