The functional equation is $f(2x+y+f(x+y))+f(xy)=yf(x)$. Note that $f$ runs from $\mathbb{R}$ to $\mathbb{R}$. We want to prove that if $f(0)=0$ then $f(x)=0$ for every real number $x$.
I have managed to prove for $x$ is an integer, since I can construct $f(x)=0$ from $f(0)=0$ by multiply for $2$ or plus. For example, $f(2\cdot(-1)+1+f(0))+f(-1)=f(-1)$, so $f(-1)=0$. Can anyone help me with $x$ is real number.
Let $P(x,y)$ be the property: $$f\big(2x + y + f(x+y)\big) + f(xy) = yf(x)$$
Looking at $P(x,-x)$ and using $f(0) = 0$, we get: $$f(x) + f(-x^2) = -xf(x) \quad\Rightarrow\quad f(-x^2) = (1+x)f(x)$$ The $x^2$ making it "symmetrical" between $x$ and $-x$, we obtain: $$f(-x^2) = (1+x)f(x) = (1-x)f(-x)$$ For $x \neq 0$, we have $1+x \neq 1-x$, and so, for $x \notin \{0,\pm1\}$, we have: $$f(x) = f(-x) \quad\Leftrightarrow\quad f(x) = 0 \tag{1}$$ As for $x = 1$: $(1+1)f(1) = (1-1)f(-1) = 0$, meaning $f(1) = 0$, and for $x = -1$, OP already proved $f(-1) = 0$, plus $f(0) = 0$ by assumption, thus $(1)$ holds for all $x \in \mathbb{R}$.
Now, $P(-x,-y)$ provides: $$f\big(-2x-y + f(-x-y)\big) + f(xy) = -yf(-x)$$ Therefore, by doing the substraction $P(x,y) - P(-x,-y)$: $$f\big(2x + y + f(x+y)\big) - f\big(-2x-y + f(-x-y)\big) = y\big(f(x) + f(-x)\big)$$
For $z \in \mathbb{R}$ such that $f(z) = 0$, $y = z - x$ in the above equation grants, thanks to $(1)$ giving $f(-z) = 0$: $$f\big(x+z\big) - f\big(-(x+z)\big) = (z-x)\big(f(x) + f(-x)\big)$$ Hence, for $x$ such that $f(x) = 0$, and using $(1)$ twice: $$f\big(x+z\big) - f\big(-(x+z)\big) = 0 \quad \Rightarrow\quad f\big(x+z\big) = f\big(-(x+z)\big) = 0$$ In particular, we have proven that (applying $(1)$ again to "change $z$ in $-z$"): $$f(x) = f(z) = 0 \,\Rightarrow\, f(x-z) = 0 \tag{2}$$
Finally, let $x \in \mathbb{R}$ be any real. From $P(x,0)$, $P(0,x)$ and $f(0) = 0$ we can see that: $$f\big(2x + f(x)\big) = f\big(x + f(x)\big) = 0$$ Which implies, by $(2)$: $$f\Big(2x + f(x) - \big(x + f(x)\big)\Big) = f(x) = 0$$ $f$ is thus indeed the zero function.
Something which I just thought about while writing this: this functional equation (with $f(0) = 0$) should still give you $f \equiv 0$ as long as you're on a commutative integral domain of characteristic $\neq 2$ I think, looking at the above proof (though I could be wrong).
I also went ahead and tried to see what happened for $f(0)$ undetermined:
With $P(0,y)$, we have: $$f\big(y + f(y)\big) = (y-1)f(0)$$ Thus, for $y = 0$: $f\big(f(0)\big) = -f(0)$.
Then, plugging this in $P(0,f(0))$ grants: $$f\Big(2 \cdot 0 + f(0) + f\big(0 + f(0)\big)\Big) + f(0 \cdot f(0)) = f\big(f(0) - f(0)\big) + f(0) = 2f(0) = f(0)^2$$ Hence: $f(0) = 0$ or $f(0) = 2$, and we've already discussed the case $f(0) = 0$.
However, the case $f(0) = 2$ seems way less easy to study, as I can't manage to see any contradiction to dismiss that case nor any useful identites to go further into it... I might come back and edit this if I find anything worthwhile. There is, like, $f\big(x(2-x)\big) = (1-x)f(x)$ from $P(x,2-x)$, and as such $f(2-x) = - f(x)$ by symmetry (and thus $f(1) = 0$), but nothing else seems to be "nice"...
Quick update ($22/09/2023$): thanks to Mohsen's kind comment below, we do have a simple way to solve the case $f(0) = 2$, by first considering $P(x-1, 1)$ to obtain that $f(2x - 1 + f(x)) = 0$, and then inputting this into $P(0, 2x-1+f(x))$, as it provides $2 = 2(2x-1+f(x))$, hence $f(x) = 2 - 2x$ after simplification.
It's then trivial to check that that function is indeed a solution, and hence the only solution of the equation when $f(0) = 2$, meaning that the problem has been solved no matter which value $f(0)$ takes.