Find $f: \mathbb{R} \to \mathbb{R}$ which satisfies $f(2x^2+2yf(z))=2xf(x)+2zf(y).$

Find $f: \mathbb{R} \to \mathbb{R}$ which satisfies $f(2x^2+2yf(z))=2xf(x)+2zf(y).$

My attempt: \begin{align} & \text{if } f \equiv 0: \text{Solution.} \\ & \text{if } f \not\equiv 0: \\ \ \\ P(x, 0, 0): & f(2x^2)=2xf(x). \tag A \label A\\ \Rightarrow & 2xf(x)=f(2x^2)=f\big(2(-x)^2\big)=-2xf(-x).\\ \text{if } x=0; & f(0)=0. \\ \text{if } x \neq 0; & f(x)=-f(-x). \tag 1 \label 1 \\ \ \\ P(0, y, z): & f(2yf(z))=2zf(y). \tag B \label B\\ \Rightarrow & f(2zf(y))=2yf(z). \\ \therefore & f^2(2yf(z))=2yf(z). \\ \therefore &\exists \ t \text{ s.t. } f(t) \neq 0.(\because f \not\equiv 0.) \\ \Rightarrow & y=\frac x {2f(t)}, z=t: \\ &f^2(x)=x. \\ \therefore & f: \text{Bijective. } \Bigg( \begin{matrix} \exists \ x' \text{ s.t. } f(x')=x \text{ for any }x. \\ f(x_1)=f(x_2) \Leftrightarrow f^2(x_1)=f^2(x_2) \Leftrightarrow x_1=x_2. \end{matrix} \Bigg) \\ \ \\ \text{By $\eqref{A} \& \eqref{B}$: } & f(2x^2+2yf(z))=f(2x^2)+f(2yf(z)). \\ \ \\ &x \leftarrow \frac x {\sqrt{2}}, y \leftarrow \frac y {2f(t)}, z \leftarrow t; \\ &f(x^2+y)=f(x^2)+f(y). \\ \ \\ \text{By } \eqref{1}: & f\big((-x^2)+(-y)\big)=f(-x^2)+f(-y). \\ y \leftarrow -y; & f(-x^2+y)=f(-x^2)+f(y). \\ \ \\ &\text{$x^2$ or $-x^2$ can represent $\forall x \in \mathbb{R}.$} \\ \therefore & f(x)+f(y)=f(x+y). \\ \end{align}

Solution 1:

Continuing in your case where $f \not \equiv 0$:

Let $t$ be any real number, and substitute $x=f(t)$ into your equation (A):

$$ f\big[2\big(f(z)\big)^2\big] = 2 f(t) f(f(t)) $$

Substitute $y=f(t), z=t$ into your equation (B):

$$ f\big[2\big(f(z)\big)^2\big] = 2 t f(f(t)) $$

So $t f(t) = t^2$. If $t \neq 0$, then $f(t) = t$. You already showed $f(0)=0$, so $f(t)=t$ for all real $t$.

The only two functions satisfying the equation are the constant zero function and the identity function.