Remark: This is a special case of Infinitydots MO 3 P6.

Let $P(x,y)$ denote substituting $x, y$ in original inequality. Take arbitrary $x>0$. By $P(x,x), 2f(x) \le x$ and by $P(x, -x), 2f(-x) - 2f(x) \le -2x$. Adding, we have $2f(-x) \le -x$ so we have that $2f(-x) = -x$. [By $P(-x, -x), 2f(-x) \ge -x$] And we also get that $2f(x) = x$ from the first two inequalities. $f(0) = 0$ can be derived seperately.

Set $x=0$ to obtain $f(0)=0$. By setting $x=y$, we have $2xf(x)\leq x^2\quad \forall x$, i.e. $f(x)\leq \tfrac{x}{2}\quad\forall x>0\ \text{and}\ f(x)\geq \tfrac{x}{2}\quad \forall x<0\qquad (\text{I})$. Assume $x<0$ and $y>0$ in the given inequality, then $$xy\geq xf(y)+yf(x)\geq xf(y)+\frac{xy}{2}$$Hence, $f(y)\geq \tfrac{y}{2}$ and by $(\text{I})$, $f(y)=\tfrac{y}{2}\quad \forall y>0$. Similarly, $f(y)=\tfrac{y}{2}\quad \forall y<0$ and therefore $\boxed{\text{S}:\ f(x)\equiv\tfrac{x}{2}}$ which obviously satisfies the given condition. $\blacksquare$

Taking $y=0$ and varying the sign on $x$ yields $f(0)=0$. Taking $y=x$ implies that $2xf(x)\le x^2$ for all $x$. Now considering $P(x,-x)$ gives $$xf(-x)-xf(x)\le -x^2\implies xf(-x)\le xf(x)-x^2\le -x^2/2.$$Multiplying by $-1$, we see $-xf(-x)\ge x^2/2$ for all $x$. This means $2xf(x)\ge x^2$ for all $x$. It follows $2xf(x)=x^2$ for all $x$, or that $\boxed{f=x/2}$.

Let $P(x,y)$ denote the given assertion. $P(x,0): xf(0)\le 0$. If $f(0)\ne 0$, we can take something the same sign as $f(0)$, which will imply $xf(0)>0$, contradiction. So $f(0)=0$. $P(x,x): 2xf(x)\le x^2\implies xf(x)\le \frac{x^2}{2}$. $P(x,-x): xf(-x)-xf(x)\le -x^2$, which implies $xf(-x)-\frac{x^2}{2}\le xf(-x)-xf(x)\le -x^2$, so $xf(-x)\le -\frac{x^2}{2}$. Swapping $x$ with $-x$ here gives $-xf(x)\le -\frac{x^2}{2}\implies xf(x)\ge \frac{x^2}{2}$ Combining with $P(x,x)$ gives $xf(x)=\frac{x^2}{2}$, so for $x\ne 0$, $f(x)=\frac{x}{2}$, and by $P(x,0)$, $f(0)=0=\frac{0}{2}$. So the only solution is $\boxed{f(x)=\frac{x}{2}}$, which works.

Note that $yf(0)\leq0$ for each $y\in\mathbb{R}$ which means $f(0)=0$. Plug $x=y$ to get $f(x)\leq\frac x2$ for every $x\geq0$ and $f(x) \geq \frac x2$ for every $x\leq 0$. Plug $x=-y\geq0$ to get $xf(-x)-xf(x)\leq-x^2$ which means $f(x)-f(-x)\geq x$ but $f(x)-f(-x)\leq \frac x2-f(-x)$. Hence, $f(-x)\leq \frac{-x}2$. From above, we had $f(x) \geq \frac x2$ for $x\leq 0$ and from here we have $f(x) \leq \frac x2$ for $x\geq 0$. Hence, equality holds and $f(x) = \frac x2$ for $x\leq 0$. Similarly $f(x) = \frac x2$ for $x\geq 0$. This gives the final answer as $f(x) = \frac x2$ for each $x$, which works.

I claim that the only solution is $\boxed{f(x) = x/2}$, and we indeed find that this solution holds. Denote by $P(x, y)$ as the assertion that $xf(y) + yf(x) \leq xy$ for all $x, y \in \mathbb{R}$. Then we consider the following: \begin{align*} P(x, x): &\implies f(x) \leq \frac{x}{2} \\ &\implies f(-x) \leq -\frac{x}{2} \\ P(-x, x): &\implies -2xf(-x) \leq x^2 \implies f(-x) \geq -\frac{x}{2}. \\ P(x, x) \wedge P(-x, x): &\implies f(-x) = -\frac{x}{2} \implies \boxed{f(x) = \frac{x}{2}}, \end{align*}as desired. $\blacksquare$

MrOrange wrote: Set $x=0$ to obtain $f(0)=0$. By setting $x=y$, we have $2xf(x)\leq x^2\quad \forall x$, i.e. $f(x)\leq \tfrac{x}{2}\quad\forall x>0\ \text{and}\ f(x)\geq \tfrac{x}{2}\quad \forall x<0\qquad (\text{I})$. Assume $x<0$ and $y>0$ in the given inequality, then $$xy\geq xf(y)+yf(x)\geq xf(y)+\frac{xy}{2}$$Hence, $f(y)\geq \tfrac{y}{2}$ and by $(\text{I})$, $f(y)=\tfrac{y}{2}\quad \forall y>0$. Similarly, $f(y)=\tfrac{y}{2}\quad \forall y<0$ and therefore $\boxed{\text{S}:\ f(x)\equiv\tfrac{x}{2}}$ which obviously satisfies the given condition. $\blacksquare$ I am not being rude, but according to your solution, I think it is the opposite of $f(y)\geq \tfrac{y}{2}$ So, it does not make any sense

$P(x, y) + P(-x, y) + P(x, -y) + P(-x, -y)$ yields \begin{align*} xf(y) + yf(x) &\leq xy \\ -xf(y) + yf(-x) &\leq -xy \\ xf(-y) - yf(x) &\leq -xy \\ -xf(-y) - yf(-x) &\leq xy \\ \end{align*}and, adding, we get $0 \leq 0$. Since the ineq is tight, $xf(y) + yf(x) = xy$. Then it's easy to get $f(0) = 0$ and $P(x, x)$ just gives the answer, $\boxed{f(x) = \frac{x}{2}}$.