Assume that
$$f(x-f(y)) \leq y f(x)+x \quad \text{ for all real } x, y . \tag{1}$$
Let $a=f(0)$. Setting $y=0$ in (1) gives $f(x-a) \leq x$ for all real $x$ and, equivalently,
$$f(y) \leq y+a \quad \text{ for all real } y . \tag{2}$$
Setting $x=f(y)$ in (1) yields in view of (2)
$$a=f(0) \leq y f(f(y))+f(y) \leq y f(f(y))+y+a .$$
This implies $0 \leq y(f(f(y))+1)$ and thus
$$f(f(y)) \geq -1 \quad \text{ for all } y>0 . \tag{3}$$
From (2) and (3) we obtain $-1 \leq f(f(y)) \leq f(y)+a$ for all $y>0$, so
$$f(y) \geq -a-1 \quad \text{ for all } y>0 \tag{4}$$
Now we show that
$$f(x) \leq 0 \quad \text{ for all real } x . \tag{5}$$
Assume the contrary, i.e. there is some $x$ such that $f(x)>0$. Take any $y$ such that
$$y<x-a \quad \text{ and } \quad y<\frac{-a-x-1}{f(x)} .$$
Then in view of (2)
$$x-f(y) \geq x-(y+a)>0$$
and with (1) and (4) we obtain
$$y f(x)+x \geq f(x-f(y)) \geq -a-1,$$
whence
$$y \geq \frac{-a-x-1}{f(x)}$$
contrary to our choice of $y$. Thereby, we have established (5).
Setting $x=0$ in (5) leads to $a=f(0) \leq 0$ and (2) then yields
$$f(x) \leq x \quad \text{ for all real } x . \tag{6}$$
Now choose $y$ such that $y>0$ and $y>-f(-1)-1$ and set $x=f(y)-1$. From (1), (5) and
(6) we obtain
$$f(-1)=f(x-f(y)) \leq y f(x)+x=y f(f(y)-1)+f(y)-1 \leq y(f(y)-1)-1 \leq -y-1,$$
i.e. $y \leq -f(-1)-1$, a contradiction to the choice of $y$.

Assume that
$$f(x-f(y)) \leq y f(x)+x \quad \text{ for all real } x, y . \tag{7}$$
Let $a=f(0)$. Setting $y=0$ in (7) gives $f(x-a) \leq x$ for all real $x$ and, equivalently,
$$f(y) \leq y+a \quad \text{ for all real } y . \tag{8}$$
Now we show that
$$f(z) \geq 0 \quad \text{ for all } z \geq 1 \tag{9}$$
Let $z \geq 1$ be fixed, set $b=f(z)$ and assume that $b<0$. Setting $x=w+b$ and $y=z$ in (7) gives
$$f(w)-z f(w+b) \leq w+b \text{ for all real } w . \tag{10}$$
Applying (10) to $w, w+b, \ldots, w+(n-1) b$, where $n=1,2, \ldots$, leads to
$$\begin{aligned} f(w)-z^{n} f(w+n b)= & (f(w)-z f(w+b))+z(f(w+b)-z f(w+2 b)) \\ & +\cdots+z^{n-1}(f(w+(n-1) b)-z f(w+n b)) \\ \leq & (w+b)+z(w+2 b)+\cdots+z^{n-1}(w+n b) \end{aligned}$$
From (8) we obtain
$$f(w+n b) \leq w+n b+a$$
and, thus, we have for all positive integers $n$
$$f(w) \leq\left(1+z+\cdots+z^{n-1}+z^{n}\right) w+\left(1+2 z+\cdots+n z^{n-1}+n z^{n}\right) b+z^{n} a . \tag{11}$$
With $w=0$ we get
$$a \leq\left(1+2 z+\cdots+n z^{n-1}+n z^{n}\right) b+a z^{n} \tag{12}$$
In view of the assumption $b<0$ we find some $n$ such that
$$a>(n b+a) z^{n} \tag{13}$$
because the right hand side tends to $-\infty$ as $n \rightarrow \infty$. Now (12) and (13) give the desired contradiction and (9) is established. In addition, we have for $z=1$ the strict inequality
$$f(1)>0 . \tag{14}$$
Indeed, assume that $f(1)=0$. Then setting $w=-1$ and $z=1$ in (11) leads to
$$f(-1) \leq-(n+1)+a$$
which is false if $n$ is sufficiently large.
To complete the proof we set $t=\min \{-a,-2 / f(1)\}$. Setting $x=1$ and $y=t$ in (7) gives
$$f(1-f(t)) \leq t f(1)+1 \leq -2+1=-1 \tag{15}$$
On the other hand, by (8) and the choice of $t$ we have $f(t) \leq t+a \leq 0$ and hence $1-f(t) \geq 1$. The inequality (9) yields
$$f(1-f(t)) \geq 0$$
which contradicts (15).