3. Let $y=t-x$. Then equation (1) can be written as
$$f(t) \leqslant t f(x)-x f(x)+f(f(x)) \text {. }$$

In equation (2), let $t=f(a), x=b$ and $t=f(b), x=a$. We get
$$\begin{array}{l} f(f(a))-f(f(b)) \leqslant f(a) f(b)-b f(b), \\ f(f(b))-f(f(a)) \leqslant f(a) f(b)-a f(a) . \end{array}$$

Adding the above two equations yields
$$\begin{array}{l} 2 f(a) f(b) \geqslant a f(a)+b f(b) . \\ \text { Let } b=2 f(a) \text {. Then } \\ 2 f(a) f(b) \geqslant a f(a)+2 f(a) f(b) \\ \Rightarrow a f(a) \leqslant 0 . \end{array}$$

Therefore, when $a>0$, $f(a) \leqslant 0$. By equation (2), for each
$$t<\frac{x f(x)-f(f(x))}{f(x)},$$

we have $f(t)<0$, which contradicts equation (3).
Thus, for all real numbers $x$,
$$f(x) \leqslant 0 \text {. }$$

Combining with equation (3), for all real numbers $x<0$,
$$f(x)=0 \text {. }$$

In equation (2), let $t=x<0$. Then
$$f(x) \leqslant f(f(x)) \text {. }$$

Hence $0 \leqslant f(0)$.
Combining with equation (4), we get $f(0)=0$.
In summary, for any $x \leqslant 0$, we have $f(x)=0$.