<s>[INST] Functions satisfying: $f(f(x)^2+f(y))=xf(x)+y$ The problem is to find all the continuous functions $f:\mathbb{R}\to \mathbb{R}$ defined by :$f(f(x)^2+f(y))=xf(x)+y$ 
I'm trying my best to figure out a way to find the expression of this unknown function by plugging some numbers, but I could not. Thanks in advance for your help. 
 [/INST] The answers are $f(x) = \pm x$.
First, plug in $x = 0$ and $y = 0$, to get that $f(f(0)^2 + f(0)) = 0$. Therefore, let $x = f(0)^2 + f(0)$. Then, we get $f(f(y)) = y$. Then, plug in $y = f(0)^2 + f(0)$. We get that $f(0) = f(0)^2 + f(0)$, or that $f(0) = 0$.
Now, let $f(a) =1$. Clearly, $f(1) =a$ as well. Now, plugging $x = a, y = 1$ in to the original equation, we get that $f(a+1) = a+1$. However, plugging in $x = 1, y = 1$ gives us that $f(a^2 + a) = a + 1$, showing that $a^2 + a = a + 1$, which means that $a = \pm 1$.
Case 1: $f(1) = 1$. Let $x = 1$. Then, we get $f(1+f(y)) = 1+y$. Now, let $y = 1$. Then, we get that $f(f(x)^2 + 1) = xf(x)+1$. However, if, in the first equation, we let $y = xf(x)$, we get that $f(1+xf(x)) = 1 + xf(x)$, which means that $f(x)^2 = x(f(x))$ or that $f(x) = x$.
Proceed similarly in the second case.
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