First assume $f(0) \ne 0$ Let $P(x,y,z)$ be assertion of this fe. $P(x,0,\frac{1}{f(0)})$ $\implies$ $f(x+1)=f(x)+1$ Let's compare $P(x,y+1,z)$ and $P(x,y,z)$ $t=x+zf(y)$ Then, $f(t+z)=f(t)+z$ Let's give $0$ to $t$.Thus,$f(x)=x+c$. Now assume $f(0)=0$ $P(0,y,z)$ $\implies$ $f(zf(y))=zf(y)$ Assume $f$ nonconstant, otherwise $f(x)=0$ indeed solution. Let $Q(y,z)$ be assertion of new fe. We can take some $f(y) \ne 0$ $Q(y,\frac{z}{f(y)})$ gives us $f(x)=x$ which is $c=0$ case So all solutions are: $\boxed{f(x)=c+x,f(x)=0}$

We claim the only such functions are $f \equiv 0$ and $f(x) = x+c$ for some real constant $c$. Clearly both these functions work. Assume $f$ to not be identically zero, and choose $y$ such that $f(y) \ne 0$. Then $zf(y)$ can take any real value $a$. Now we get $f(x+a) = f(x) + a = f(a) + x$, so $f(x) - x$ is a constant, say $c$. This gives us $f(x) = x + c$. Therefore, we have obtained both solutions. $\square$

Answer : $ f(x) = x + c $ and $ f(x) = 0 $. Solution : Let $ P(x,y,z) $ denote the given assertion. Case 1 : $ f(x) = 0 $, which is true. Case 2 : $ f(x) \ne 0 $. $ P(0,y,\frac {z} {f(y)}) \Rightarrow f(x) = x + c $. (for some constant c) If we insert this into the equation, we will see that is correct.

The answers are : $f_1(x)=0,(\forall)x\in\mathbb{R}$ or $f_2(x)=x+c,(\forall)x\in\mathbb{R}, c\in\mathbb{R}.$ It's easy to see that these functions satisfy the given functional equation. We will prove these are the only ones. Denote by $\mathcal{P}(x,y,z):$$$f(x+zf(y))=f(x)+zf(y),(\forall)x,y,z\in\mathbb{R}.$$Supose $f\not\equiv 0,$ then there exists some $y_0\in\mathbb{R}$ such that $f(y_0)\neq0.$ Denote $f(0)$ by $c$. $\mathcal{P}(x,y_0,-\frac{x}{f(y_0)})$ gives : $c=f(x)-x$ which is equivalent to $f(x)=x+c,(\forall)x\in\mathbb{R},c\in\mathbb{R}.$ Thus, $f_1$ and $f_2$ are the only solutions$_{\blacksquare}.$

Clearly $f \equiv 0$ is a solution. Henceforth assume that there exist $x \in \mathbb{R}$ such that $f(x) \neq 0$. Choose $y$ such that $f(y) \neq 0$. $P(0,y,\frac{a}{f(y)}) \implies f(a)=f(0)+a$. It follows that $f(x)=x+c$ for all $x$ for some constant $c$. We may easily check that all functions of this form are solutions. $\therefore f \equiv 0$ or $f(x) = x+c$ for some constant $c \in \mathbb{R}$.

The only solutions are $f\equiv 0$ and $f(x) = x + c$ for some constant $c$, which obviously work. Now we show they are the only solutions. Suppose that $f$ wasn't identically zero. Note that $zf(y)$ is surjective over reals, so we have \[ f(x+y) = f(x) + y \forall x,y \in \mathbb R\] Setting $x = 0$ gives $f(y) = y + f(0) \forall y \in \mathbb R$, and we are done by setting $c = f(0)$.

now it's obvious that $ f(x)=0 $ is one of the our solutions then let's search for others Let $P(x,y,z)$ be assertion from our function we can see that it's surjective Now there exist some u that $f(u)\neq 0$ now let's look $P(x,u,\frac{x}{f(u)}) \rightarrow f(2x)=f(x)+x $ now we can se that $f(2x)-f(x)=x$ which means that it's const now for some c we can say that $f(x)=x+c$ and the solution is $\boxed{f(x)=0} \land \boxed{f(x)=x+c}$

First of all, it's obvious that $f(x) \equiv 0$ is a solution since $$ 0=0+0 $$Let us assume, from now on, that $f(x) \not \equiv 0$; that is, there exists a number $x_0$ such that $f(x_0)=y_0 \neq 0$. Substituting in the original expression for $x=0, y=x_0, z$, we get $$ f(0+zf(x_0)) = zf(x_0) + f(0)$$Since we can make $z$ take on any value, we can make $zf(x_0)$ take on any value and therefore $f(x)=x+f(0)$ or we have the final solutions $f(x) \equiv 0$ or $f(x) \equiv x+c$ for any constant $c$.