Taking $g(x)=x-f(x)$ we’ve $g(g(x))=g^2(x)=g(x)-\frac {x}{2}$ so $g$ is one to one. Now continuity of $f$ implies $g$ is continuous as well. As because it’s Onto so it’s a monotonic function too. Now from $g^4(x)=-\frac {x}{4}$ we get $g$ is decreasing function. So $x>y\implies g(x)<g(y)\implies g^2(x)>g^2(y)\implies g(x)-\frac {x}{2}>g(y)-\frac {y}{2}$ and that implies now $y>x$. So $g$ must be constant, but that’s impossible either, so no such $f$ exist.

More involved approach, working mostly on $f(\cdot)$: First, notice that $f(\cdot)$ is surjective (for any real $r$, taking $t=2r-f(2r)$ yields $f(t)=r$). Now, let us continue with a few moves on the assertion: $\star$ Plugging $x\mapsto x-f(x)$ yields, $$ f(x-f(x)-f(x-f(x)))=\frac{x-f(x)}{2}\implies f\left(\frac{x}{2}-f(x)\right)=\frac{x-f(x)}{2}. $$$\star$ Further plugging $x\mapsto x-f(x)$, in the previous item, yields, $$ f\left(\frac{x-f(x)}{2}-f(x-f(x))\right)=\frac{x-f(x)-f(x-f(x))}{2} \implies f\left(-\frac{f(x)}{2}\right)=\frac{x}{4}-\frac{f(x)}{2}. $$From this item, we deduce that $f(\cdot)$ is injective. Since $f(\cdot)$ is both surjective and injective, and is continuous, it must be monotonic. $\star$ Plugging $x\mapsto x-f(x)$ one more time, in the previous item, we have, $$ f\left(-\frac{f(x-f(x))}{2}\right)=\frac{x-f(x)}{4}-\frac{f(x-f(x))}{2}\implies f\left(-\frac{x}{4}\right)=-\frac{f(x)}{4}. $$Hence, $f(0)=0$. Now, we will continue by proving that $f(\cdot)$ must be increasing. Suppose not, and $x<y \implies f(x)>f(y)$. Then, $x-f(x)<y-f(y)$, hence, $$ \frac{x}{2}=f(x-f(x))>f(y-f(y))=\frac{y}{2}\implies x>y, $$a contradiction. Hence, $f(x)>0$ for $x>0$, and $f(x)<0$, for $x<0$. Next, we claim that the only fixed point of $f(\cdot)$ is 0. Suppose again that there is a $y\neq 0$ such that $f(y)=y$. Then, $$ 0=f(0)=f(y-f(y))=\frac{y}{2}, $$again, a contradiction. Hence, on $(0,\infty)$, $x-f(x)$ should not switch signs (otherwise, due to continuity, it would have a fixed point). Now, if $f(x)>x$ on $(0,\infty)$, then $x-f(x)$ is a negative number, however, $f(x-f(x))=x/2>0$, again, a contradiction. Hence finally, $f(x)<x$ on $(0,\infty)$. In this case, since $x-f(x)>0$, we have, $$ \frac{x}{2}=f(x-f(x))<x-f(x)\implies f(x)<\frac{x}{2}, $$and therefore, $$ \frac{x}{2}=f(x-f(x))<\frac{x-f(x)}{2}\implies f(x)<0, $$a contradiction. Hence, no such $f(\cdot)$ exists.

subham1729 wrote: Taking $g(x)=x-f(x)$ we’ve $g(g(x))=g^2(x)=g(x)-\frac {x}{2}$ so $g$ is one to one. Now continuity of $f$ implies $g$ is continuous as well. As because it’s Onto so it’s a monotonic function too. Now from $g^4(x)=-\frac {x}{4}$ we get $g$ is decreasing function. So $x>y\implies g(x)<g(y)\implies g^2(x)>g^2(y)\implies g(x)-\frac {x}{2}>g(y)-\frac {y}{2}$ and that implies now $y>x$. So $g$ must be constant, but that’s impossible either, so no such $f$ exist. why g is 1 to 1?