edit: yup this is wrong but it seems like it's in the right direction strange as my solutions for $f(1) = 2$ are $\frac{1}{x}\cdot \text{what they should be}$ (which is why i ruled them out bc they dont satisfy fe)

parmenides51 wrote: Find all functions $f: R \to R$ that satisfy $$f (x^2y) + 2f (y^2) =(x^2 + f (y)) \cdot f (y)$$for all $x, y \in R$ Let $P(x,y)$ be the assertion $f(x^2y)+2f(y^2)=(x^2+f(y))f(y)$ $P(x,0)$ $\implies$ $f(0)=0$ $P(1,x)$ $\implies$ new assertion $Q(x)$ : $2f(x^2)=f(x)^2$ Subtracting $P(1,y)$ from $P(x,y)$, we get new assertion $R(x,y)$ : $f(x^2y)=x^2f(y)$ $Q(1)$ $\implies$ $f(1)\in\{0,2\}$ If $f(1)=0$, $R(x,1)$ $\implies$ $f(x^2)0$ and then $Q(x)$ implies $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ Which indeed fits If $f(1)=2$, $R(x,1)$ $\implies$ $f(x^2)=2x^2$ $Q(-1)$ $\implies$ $f(-1)\in\{-2,2\}$ If $f(-1)=2$, $R(x,-1)$ $\implies$ $f(-x^2)=2x^2$ and so $\boxed{\text{S2 : }f(x)=2|x|\quad\forall x}$ Which indeed fits If $f(-1)=-2$, $R(x,-1)$ $\implies$ $f(-x^2)=-2x^2$ and so $\boxed{\text{S3 : }f(x)=2x\quad\forall x}$ Which indeed fits

I believe this is right, but I'm a relative beginner to functional equations, so it might be wrong. Edit: Sniped twice! The solutions are $f \equiv 0, f(x)=2x, f(x)=2|x|$. Let $P(x,y)$ denote the given assertion. By $P(x,0)$ we get $3f(0)=f(0)^2+x^2f(0)$. It follows that we must have $f(0)=0$. From $P(x,1)$ we get $f(x^2)+2f(1)=x^2f(1)+f(1)^2$. Letting $f(1)=c$ we get $f(x^2)=cx^2+c^2-2c$. Substituting $k=x^2$ where $k$ is a nonnegative real, we have $f(k)=ck+c^2-2c$. However taking $k=0$ we have $f(0)=c^2-2c=0$, so $c^2-2c=0 \implies c=0,2$. Thus for nonnegative $k$, we have $f \equiv 0$ or $f(k)=2k$. Now we must consider what happens for negative inputs. Case 1: For positive inputs $f \equiv 0$. From $P(1,-y)$ where $y \geq 0$ we get $f(-y)+2f(y^2)=f(-y)+f(-y)^2 \implies f(-y)^2=0 \implies f(-y)=0$, so $f \equiv 0$ for negative inputs as well. This gives us $f \equiv 0$. It is easy to see that this works. Case 2: For positive inputs $f(x)=2x$. From $P(1,-y)$ where $y \geq 0$, we get $f(-y)+2f(y^2)=f(-y)+f(-y)^2 \implies f(-y)^2=4y^2$. Thus $f(-y)=\pm 2y$. Thus, we have $f(x)=2x,-2x$ for $x < 0$. If $f(x)=2x$ for all $x$, the original condition holds. Now we have to look at: \[ f(x)= \begin{cases} 2x & x \geq 0\\ -2x & x <0 \end{cases} \]Which is just $f(x)=2|x|$. The condition then becomes $2|x^2y|+4|y^2|=2x^2|y|+4y^2 \implies x^2|y|+2y^2=x^2|y|+2y^2$, so this works as well. We must deal with the pointwise trap. Suppose there exist positive reals $a,b$ such that $f(-a)=2a$ and $f(-b)=-2b$, and $a \neq b$. Write $b=ak^2$ for positive $k$. Then by $P(k,-a)$ we get $f(-b)+2f(a^2)=k^2f(-a)+f(-a)^2 \implies -2b+4a^2=2b+4a^2 \implies b=-b \implies b=0$, but $b$ is positive so we have reached a contradiction. Thus we have eliminated the pointwise trap, so there are no more solutions. Thus we are done. $\blacksquare$

Let $P(x,y)$ be the assertion of $f (x^2y) + 2f (y^2) =(x^2 + f (y)) \cdot f (y)$. $P(x,0)\implies 3f(0)=f(0)\cdot (x^2+f(0))\implies f(0)=0\text{ or } x^2=3-f(0)$, thus $f(0)=0$. $P(0,y)\implies 2f (y^2) =f (y)^2$ $2f(1)=f(1)^2\implies f(1)=2 \text{ or } f(1)=0$ $P(x,1)\implies f(x^2)+2f (1)=(x^2+f(1))f(1)\implies \frac{f(x)^2}{2}+2f(1)=x^2f(1)+f(1)^2$. $f(x)^2=0\implies f(x)=0$. This does work. $f(x)^2=4x^2\implies f(x)=\pm 2x$. $f(x^2y^2)=x^2f(y^2)=y^2f(x^2) \implies f(x)=cx\,\forall x\in \mathbb R^+$. Thus,$ f(y)^2=2cy^2$, if $y$ is positive, then we get that $c=2$, thus $f(x)=2x\,\forall x\in \mathbb R^+$. This does work. Now if $y$ is negative, we get that $f (x^2y) + 4y^2=(x^2 + f (y)) \cdot f (y)$, we see that $f(y)=2y \text{ or } -2y$. This does work. Hence, we have: $\boxed{f(x)=2x\,\forall x\in \mathbb R_{\geq 0} \text{ and } f(x)=2x \text{ or }-2x\,\forall x\in \mathbb R^- }$ and $\boxed{f(x)=0\forall x\in \mathbb R}$

Let $P(x, y)$ denote the given assertion. We claim that the only solutions are $\boxed{f \equiv 0, f(x) = 2x, f(x) = 2|x|}$. It is easy to verify that these solutions work. We now prove that these are the only solutions. $P(0, 0) \implies 3f(0) = [f(0)]^2 \implies f(0) \in \{0, 3 \} $. If $y = 0$, the latter is clearly not true for all choices of $x$, so $f(0) = 0$. $P(0, 1) \implies f(0) + 2f(1) = [f(1)]^2 \implies 2f(1) = [f(1)]^2 \implies f(1) \in \{0, 2\}$. Case 1: Assume that $f(1) = 0$. $P(x, 1) \implies f(x^2) = 0$. $P(0, -1) \implies f(-1) = 0$. So, $P(x, -1) \implies f(-x^2) = 0$. Thus, $f(x^2) = 0 \wedge f(-x^2) = 0 \implies \boxed{f(x) = 0}$. Case 2: Assume that $f(1) = 2$. $P(x, 1) \implies f(x^2) = 2x^2$. $P(0, -1) \implies f(-1) = \pm 2$. Subcase 1: If $f(-1) = -2$, then $P(x, -1) \implies f(-x^2) = -2x^2$. Then, $f(x^2) = 2x^2 \wedge f(-x^2) = -2x^2 \implies \boxed{f(x) = 2x}$. Subcase 2: If $f(-1) = 2$, and $P(x, -1) \implies f(-x^2) = 2x^2$. Then, $f(x^2) = 2x^2 \wedge f(-x^2) = 2x^2 \implies \boxed{f(x) = 2|x|}$. $\blacksquare $

I just realized I forgot about the pointwise trap in my solution. It can be repaired as such: The solutions are $f \equiv 0, f(x)=2x, f(x)=2|x|$. We must deal with the pointwise trap. Suppose there exist positive reals $a,b$ such that $f(-a)=2a$ and $f(-b)=-2b$, and $a \neq b$. Write $b=ak^2$ for positive $k$. Then by $P(k,-a)$ we get $f(-b)+2f(a^2)=k^2f(-a)+f(-a)^2 \implies -2b+4a^2=2b+4a^2 \implies b=-b \implies b=0$, but $b$ is positive so we have reached a contradiction. Thus we have eliminated the pointwise trap, so there are no more solutions. Thus we are done. $\blacksquare$