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Let $a=f(0)$. Setting $x=y=0$ in the given equation gives $f(a)=a^2$. Setting $y=0$ gives $$f(x^2+a)=(f(x){)}^2,\forall x\in \mathbb{R}.(\circ )$$ Setting $x=a$ gives $f(a^2+a)=(f(a){)}^2=a^4$. Assume that $a<0$. Then there exists $b>0$ such that $a=-b^2$. Setting $x=b$ in (\circ) gives $a=f(0)=f(b^2+a)=(f(b){)}^2\ge 0$, a contradiction. Hence $a\ge 0$. Setting $x=\sqrt{a}$, $y=a$ in the given equation gives $f(a+a^2)=(f(\sqrt{a})+a^2{)}^2$. We have obtained $f(a+a^2)=(f(\sqrt{a})+a^2{)}^2=a^4$, i.e. $f(\sqrt{a})(f(\sqrt{a})+2a^2)=0$. Assume $f(\sqrt{a})=-2a^2$. Setting $x=\sqrt{\sqrt{a}+2a^2}$, $y=\sqrt{a}$ gives $$f(x^2+f(y))=f(\sqrt{a}+2a^2-2a^2)=f(\sqrt{a})=-2a^2<0.$$ But the given equation gives $f(x^2+f(y))=(f(x)+y^2{)}^2\ge 0$, a contradiction. Hence $f(\sqrt{a})=0$. Setting $x=0$ in the given equation gives $f(f(y))=(a+y^2{)}^2,\forall y\in \mathbb{R}$. Setting $y=\sqrt{a}$ gives $a=f(0)=f(f(\sqrt{a}))=(a+a{)}^2=4a^2$. We have two cases: $a=\frac{1}{4}$ or $a=0$. Assume $a=\frac{1}{4}$. Then $f(\frac{1}{2})=0$. Because $f(f(y))=(a+y^2{)}^2$, we have $f(\frac{1}{4})=\frac{1}{16}$. If $x$ and $y$ are real numbers such that $x^2+f(y)=\frac{1}{2}$, the given equation gives $f(x)=-y^2$. This is true for, e.g., $x=\frac{\sqrt{7}}{4}$ and $y=\frac{1}{4}$ and we conclude that $f(\frac{\sqrt{7}}{4})=-\frac{1}{16}<0$. But setting $x=\sqrt{\frac{\sqrt{7}-1}{4}}$ in (\circ) gives $f(\frac{\sqrt{7}}{4})\ge 0$, a contradiction. Hence $f(0)=a=0$ and we can simplify obtained identities to $f(x^2)=(f(x){)}^2$ and $f(f(y))=y^4$. From the first identity we have $f(x)\ge 0$ for all $x\ge 0$. Assume there is $t>0$ such that $f(t)<t^2$. Setting $x=\sqrt{t^2-f(t)}$ and $y=t$ in the given equation gives $$f(t^2)={\left(f(\sqrt{t^2-f(t)})+t^2\right)}^2\ge t^4,$$ therefore $(f(t){)}^2\ge t^4$ which is a contradiction with the assumption $f(t)<t^2$. Hence for all $x\ge 0$ holds $f(x)\ge x^2$. This implies that $x^4=f(f(x))\ge (f(x){)}^2\ge x^4$ for all $x>0$, which is possible only if $f(x)=x^2$ for all $x\ge 0$. Let $w>0$. Setting $x=-w$ in $f(x^2)=(f(x){)}^2$ gives $f(-w)=w^2$ or $f(-w)=-w^2$. But if $f(-w)=-w^2$, setting $x=w$, $y=-w$ in the given equation gives $0=f(0)=f(w^2-w^2)=f(x^2+f(y))=(f(x)+y^2{)}^2=4w^4>0$, a contradiction. Hence the only possible solution is $f(x)=x^2,\forall x\in \mathbb{R}$. We check directly that it really satisfies the given equation.