The answer is $f(r)=2^{-\nu_2(r)}$ for $r\neq 0$ and $f(0)=0$. It is easy to check that this satisfies the conditions. The second and third are trivial; as for the first, we know that $$\nu_2(x+y)\geq \min\{\nu_2(x),\nu_2(y)\}$$so $$f(x+y)=2^{-\nu_2(x+y)}\leq 2^{-\min\{\nu_2(x),\nu_2(y)\}}\leq 2^{-\nu_2(x)}+2^{-\nu_2(y)}=f(x)+f(y)$$ On to the proof that this is the only function that satisfies the conditions. First, from the second condition, by considering the pairs $(x,y)=(0,2),(1,2)$, we get $f(0)=0,f(1)=1$. We also get $f(2^k)=\frac{1}{2^k}$. In fact, $f$ is multiplicative under $\mathbb{Q}$. Furthermore, we may obtain $f(-1)=1$ from $(x,y)=(-1,-1)$, and $f(-r)=f(r)$ for all rational numbers $r$. From the first condition, we get $f(y)\geq |f(x+y)-f(x)|$ for all pairs $(x,y)$. Set $y=2^a$. We get $\frac{1}{2^a}\geq |f(x+2^a)-f(x)|$. Choose an arbitrary positive integer $t$ and write it in the form $2^{a_1}+2^{a_2}+\cdots+2^{a_n}$. Using the triangle inequality, we obtain $$|f(x+t)-f(x)|\leq |f(x+t)-f(x+t-2^{a_n})|+|f(x+t)-f(x+t-2^{a_n}-2^{a_{n-1}})|+\cdots+|f(x+2^{a_1})-f(x)|\leq \frac{1}{2^{a_n}}+\cdots+\frac{1}{2^{a_1}}<\frac{2}{2^{a_1}}$$ We now claim that for any prime $p\geq 3$, we have $f(p)=1$, which automatically implies the desired conclusion. Let $t=p^u-p^v$ and $x=p^v$. Then we have $$\frac{2}{2^{\nu_2(p^u-p^v)}}\geq |f(p)^u-f(p)^v|$$If $f(p)>1$, then we fix $u-v$ and let $u$ be large. This yields a contradiction. If $f(p)<1$, let $x=1$ and $t=p^k-1$. We have $$\frac{2}{2^{\nu_2(p^k-1)}}\geq |1-f(p)^k|$$Setting $k$ to be a large power of 2, we get a contradiction. Hence, $f(p)=1$ and we are done.

Why is this true ? ACGNmath wrote: From the first condition, we get $f(y)\geq |f(x+y)-f(x)|$ for all pairs $(x,y)$.

Nofancyname wrote: Why is this true ? ACGNmath wrote: From the first condition, we get $f(y)\geq |f(x+y)-f(x)|$ for all pairs $(x,y)$. First, condition (i) implies $f(y) \geq f(x+y) - f(x)$. Moreover, since we already know that $f(y)=f(-y)$ for all $y$, we obtain $$f(y) + f(x+y) = f(-y) + f(x+y) \geq f(x) \Longrightarrow f(y)\geq f(x) - f(x+y).$$

The answer is $f(x)\equiv 2^{-\nu_2(x)}$, f(0)=0. By induction, we obtain $f(2^k)=2^{-k}$ for all $k\in \mathbb{Z}$. Furthermore, $f(-1)=1$ Since any integer has a binary representation, say $n=\sum\limits_{k=1}^t 2^{a_k}$, then $f(n) = f(\sum\limits_{k=1}^t 2^{a_k}) \le \sum\limits_{k=1}^t f(2^{a_k}) = \sum\limits_{k=1}^t 2^{-a_k} < 2$. Thus, $f(x)<2$ for all $x\in \mathbb{N}$. Furthermore, $f(x)^n=f(x^n)<2$ so $f(x)\le 1$ or we can choose suitable $n$ to get a contradiction. Now, I claim $f(x)=1$ for all odd $x$, which solves the problem. Note $1-f(x)^{2^n} = 1-f(x^{2^n}) <f(1-x^{2^n}) = f(x^{2^n}-1)$. By LTE, $\nu_2(x^{2^n}-1)\ge n+2$. Therefore, $1-f(x)^{2^n} < 2^{-n-1}$ by writing $x^{2^n}-1$ in binary. It follows that for all $x$ odd, $f(x)=1$, so $f(x)\equiv 2^{-\nu_2(x)}$