proximo wrote: Find all pairs $(f,h)$ of functions $f,h:\mathbb{R}\rightarrow\mathbb{R}$ such as for all real $x$ and $y$ the equation holds. $$f(x^2+yh(x))=xh(x)+f(xy)$$ Let $P(x,y)$ be the assertion $f(x^2+yh(x))=xh(x)+f(xy)$ Let $a=f(0)$ Let $x\ne 0$ : If $h(x)\ne x$ : Subtracting $P(x,0)$ from $P(x,\frac{x^2}{ x-h(x)})$, we get $f(x^2)=a$ If $h(x)=x$ : $P(x,0)$ $\implies$ $f(x^2)=x^2+a$ So $\forall x\ne 0$, either $f(x^2)=a$, either $f(x^2)=x^2+a$ Let $x\ne 0$ : $P(x,0)$ $\implies$ $f(x^2)=xh(x)+a$ and so : Either $a=xh(x)+a$ and so $h(x)=0$ Either $x^2+a=xh(x)+a$ and so $h(x)=x$ If $h(t)=0$ for some $t\ne 0$, then $P(t,\frac xt)$ $\implies$ $f(x)=f(t^2)$ constant and $P(x,y)$ becomes $h(x)=0$ $\forall x\ne 0$ Hence the solution : $\boxed{\text{S1 : }f(x)=a\text{ }\forall x\text{ and }h(x)=0\text{ }\forall x\ne 0\text{ and }h(0)=b}$ which indeed is a solution, whatever are $a,b\in\mathbb R$ If $h(0)\ne 0$, then $P(0,\frac x{h(0)})$ $\implies$ $f(x)=a$ constant and so the same solution than just above. If $h(0)=0$ and $h(x)\ne 0$ $\forall x\ne 0$, then $h(x)=x$ $\forall x$ and $P(x,y)$ becomes new assertion $Q(x,y)$ : $f(x^2+xy)=x^2+f(xy)$ $Q(x,0)$ $\implies$ $f(x^2)=x^2+a$ $Q(x,-x)$ $\implies$ $f(-x^2)=-x^2+a$ And so $f(x)=x+a$ $\forall x$ and the solution : $\boxed{\text{S2 : }f(x)=x+a\text{ }\forall x\text{ and }h(x)=x\text{ }\forall x}$; which indeed is a solution, whatever is $a\in\mathbb R$

Instead of $h$, use $g$ as the function. Let $P(x,y)$ be the assertion $f(x^2+yg(x))=xg(x)+f(xy)$. If $g(0)\ne0$: $P\left(0,\frac x{g(0)}\right)\Rightarrow f(x)=f(0)$ $P(x,y)\Rightarrow xg(x)=0$, hence the solution $\boxed{(f(x),g(x))=\left(c,\begin{cases}0&\text{if }x\ne0\\d&\text{if }x=0\end{cases}\right)}$ for $c,d\in\mathbb R$ which is a solution. Otherwise, assume $g(0)=0$. If $g(x)=x$ for all $x$, then $f(x^2+xy)=x^2+f(xy)$. $P(x,0)\Rightarrow f(x^2)=x^2+f(0)$ $P(x,-x)\Rightarrow f(-x^2)=-x^2+f(0)$ Thus $\boxed{(f(x),g(x))=(x+c,x)}$. Assume $g(j)\ne0$ for some $j$. Note that $j\ne0$. $P\left(j,\frac{j^2}{j-g(j)}\right)\Rightarrow g(j)=0$ $P\left(j,\frac xj\right)\Rightarrow f(x)=f(j^2)\Rightarrow f(x)=0$, leading to the first solution.

Let $P(x,y)$ be the above assertion. Then, $P(0,y)$ gives $$ f(y\cdot h(0))=f(0) \quad \forall \, y\in \mathbb R. \qquad (\star) $$Let's suppose $h(0)\neq 0$. Then, I claim that $f$ is constant, $h(x)=0$ for every non-zero real, and $h(0)$ can take any real value.

Now, let's suppose $h(0)=0$. We will distinguish two cases: Case 1. There exists a real $\alpha\neq 0$ satisfying $f(\alpha)=0$. I claim that the only solutions for Case 1 are $f$ constant, $h\equiv 0$.

Case 2. There doesn't exist any $\alpha\neq 0$ satisfying $h(\alpha)=0$. I claim that the only solution to this case will be $f(x)=x+c$ ; $h(x)=x$ for every real $x$.

The three solutions can be grouped as $$ \boxed{ \left\{\begin{array}{lll} f \thickspace \text{constant,} \thickspace h(x)=0 \quad \forall x\in \mathbb R \setminus \{0\} \\ f(x)=x+c \thickspace ; \thickspace h(x)=x \quad \forall x\in \mathbb R. \\ \end{array} \right. } $$