$$ f:\mathbb{R} \to \mathbb{R}, f(xf(x)+2y)=f(x^2)+f(y)+x+y-1 $$ Those are my attempt. $$ P(0, y): f(2y) = f(0)+f(y)+y-1 \\ y=0; \ f(0)=2f(0)-1 \implies f(0)=1. \\ \ \\ P(x, 0): f(xf(x))=f(x^2)+x \\ x=1; \ f(f(1))=f(1)-1.\\ \ \\ P(1, f(1)): f(3f(1))=f(f(1))+2f(1) = 3f(1)-1. \\ P(1, 3f(1)): f(7f(1))=f(3f(1))+4f(1) = 7f(1)-1. \\ \cdot \\ \cdot \\ \cdot \\ f((2^n-1)f(1))=(2^n-1)f(1)-1. $$
Please find all the functions that are possible with full process.
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I can help to find only continuous in $0$ function that satisfy this equation. Let $f:\mathbb R\to\mathbb R$ be a continuous in $x=0$ function such that: $$f(xf(x)+2y)=f(x^2)+f(y)+x+y-1 \quad\quad (1)$$ If we substitute to (1) $x\leftarrow 0$ and $y\leftarrow 0$ then we have $$ f(0)=f(0)+f(0) - 1\\ f(0)=1\quad\quad (2) $$
If we substitute to (1) $x\leftarrow 0$ then we have $$ f(2y)=f(0)+f(y) +0 +y -1\\ f(2y)=1+f(y) +0 +y -1\\ f(2y)= f(y)+y\quad\quad (3) $$ Let $x\in\mathbb R$, then according to (3) we have: $$ \begin{align*} f(x) &= f\left(2\cdot\frac{1}{2}x\right)=f\left(\frac{1}{2}x\right)+\frac{1}{2}x = f\left(2\cdot\frac{1}{4}x\right)+\frac{1}{2}x =\\ &=f\left(\frac{1}{4}x\right)+\frac{1}{4}x+\frac{1}{2}x = \cdots = \\ &=f\left(\frac{1}{2^n}x\right) + \sum_{i=1}^{n}\frac{1}{2^i}x \end{align*}, $$ but (thanks that $f$ is continuous in $0$) $$ \lim_{n\to\infty} f\left(\frac{1}{2^n}x\right) = f(0) = 1 $$ and $$ \lim_{n\to\infty} \sum_{i=1}^{n}\frac{1}{2^i}x = x\cdot\lim_{n\to\infty} \sum_{i=1}^{n}\frac{1}{2^i} = x\cdot 1 = x. $$ Finally $$ x+1 = \lim_{n\to\infty} \sum_{i=1}^{n}\frac{1}{2^i}x + \lim_{n\to\infty} f\left(\frac{1}{2^n}x\right)= \lim_{n\to\infty}\left(f\left(\frac{1}{2^n}x\right) + \sum_{i=1}^{n}\frac{1}{2^i}\right) = \lim_{n\to\infty}f(x)=f(x). $$ It's easy to check that $f(x)=x+1$ satisfay (1), but I don't know what about functions which aren't continuous in $x=0$. In fact if $f$ has limit $g$ in $x=0$, then - using (1) - we can show that $$\lim_{x\to 0}f(x)=1=f(0), $$ becouse $f(xf(x) + 2x)\to g$ and $f(x^2) + f(x) + 2x - 1 \to 2g -1$, so $$g = 2g -1\\ g = 1.$$
I am quite proud of this solution. For the sake of being self-contained, the functional equation we are tasked to solve is $$ f(xf(x) + 2y) = f(x^2) + f(y) + x + y - 1. $$ Plugging in $x = y = 0$ gives us that $$ f(0) = 2f(0) - 1 \quad \text{and thus} \quad f(0) = 1. $$ Plugging in $x = 0$ then gives us that $$ f(2y) = f(y) + y, \qquad (*) $$ so we may rewrite our functional equation to $$ f(xf(x) + 2y) = f(x^2) + f(2y) + x - 1. $$ Plugging in $y = 0$ gives us that $$ f(xf(x)) = f(x^2) + x \qquad (**) $$ and so we may further rewrite our functional equation to $$ f(xf(x) + 2y) = f(xf(x)) + f(2y) - 1. $$
We plug in $y = -xf(x)$. We find that $$ f(-xf(x)) = f(xf(x)) + f(-2xf(x)) - 1 = f(xf(x)) + f(-xf(x)) - xf(x) - 1, $$ where we used $(*)$. We may rewrite this to $$ f(xf(x)) = xf(x) + 1. $$ This means that on the set $A = \{ xf(x) \mid x \in \mathbb{R} \}$, the function $f$ is identically equal to $x+1$.
Our functional equation can thus further be rewritten to $$ f(xf(x) + 2y) = xf(x) + f(2y), \qquad (***) $$ Note that we may now also simplify $(**)$ to $$ xf(x) + 1 = f(x^2) + x. $$ Plugging in $-x$ for $x$ here gives us that $$ -xf(-x) + 1 = f(x^2) - x, $$ and so subtracting these two, we find that $$ xf(x) - ( -xf(-x) ) = 2x. $$ This means that for any $x \in \mathbb{R}$, there exist $a,a' \in A$ with $a - a' = 2x$. There is no necessity for this additional notation, but it will make the final stretch less messy. For any $x \neq 0$, we may divide by it to also find that $$ f(x) + f(-x) = 2. $$ Note that this also holds for $x = 0$, so it holds for all $x \in \mathbb{R}$.
Set $B = \{ x \in \mathbb{R} \mid f(x) = x+1 \}$. We show that $B = \mathbb{R}$. We have seen that $A \subset B$ and the above shows that also $-A \subset B$. Now, let $x \in \mathbb{R}$ be arbitrary and write $2x = a - a'$ for some $a, a' \in A$ as above. Set $xf(x) = a$ and $2y = -a'$ in $(***)$. It gives us that $$ f(2x) = f(a + (-a')) = a + f(-a') = a - a' + 1 = 2x+1. $$ Since $x \in \mathbb{R}$ was arbitrary, we are done.