Determine all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$(x-2)f(y)+f\big(y+2f(x)\big)=f\big(x+yf(x)\big)$$ for all $x,y\in\mathbb{R}$.
Source: Math Excalibur Volume 22, Number 1 Problem 522
(Functional equations are on-topic, there are few previous functional equations by other users.)
All such function(s) is(are):
$f\equiv 0$ and $f(x)=x-1$. These can be easily seen to work via some algebra.
How so?
Suppose $f(0)=0$ at first. Putting $x=0$ gives $-2f(y)+f(y)=0$, so $f(y)=0$ for all $y$, i.e., $f$ is the all-zero function.
The other case is
when $f(0)\ne 0$. Now plugging $y=0$ gives $(x-2)f(0)=-f(2f(x))+f(x)$. Note that if $f(x_1)=f(x_2)$, then $-f(2f(x_1))+f(x_1)=-f(2f(x_2))+f(x_2)$, so the last relation implies $(x_1-2)f(0)=(x_2-2)f(0)$, so $x_1=x_2$. This means $f$ is injective.
Next, plug in
$x=2$ and $y=0$, to get $f(2f(2))=f(2)\implies 2f(2)=2$ (because of injectivity). Thus $f(2)=1$, and by injectivity, $2$ is the only number whose image is $1$.
We will derive one last thing before the finale.
We claim that the only possible preimage of $0$ is $1$. Indeed, if $f(c)=0$, then putting $y=2,x=c$ gives $(c-2)\cdot 1+1=0$, which gives $c=1$.
Now for the finish. We plug in
$y=\frac{2f(x)-x}{f(x)-1}$ (for $x\ne 2$). Note that this magical substitution makes $f(y+2f(x))$ and $f(x+yf(x))$ cancel out, so we are left with (using $x\ne 2$)$$(x-2)f\left(\frac{2f(x)-x}{f(x)-1}\right)=0\implies f\left(\frac{2f(x)-x}{f(x)-1}\right)=0.$$But by the last inference, this must mean $\frac{2f(x)-x}{f(x)-1}=1$, which can be rearranged to $f(x)=x-1$. We derived this assuming $x\ne 2$, but this is true for $x=2$ as well, so in fact $f(x)=x-1$ has to hold for all $x$.
