My approach. Anyone can finish? $\boxed{\text{Answer. }f(x)=x.}$ Easily we can make induction (using $(a)$ and $(b)$) and get that $f(x)=x$ for all $\mathbb{Z}$. From $c$ we obtain that $f\left(\frac{1}{x}\right) =\frac{1}{x}$ for all $x\in \mathbb{Z}$, thus again by induction with $(b)$, we get that $f(x)=x \forall x\in \mathbb{Q}$.

Replace $x,y$ in (b) with $\frac{1}{x},\frac{1}{y}$, then using (c) we can rewrite as $$f(\frac{xy}{x+y})=\frac{f(x)y^2+f(y)x^2}{(x+y)^2}.$$Replacing $y$ by 1 and then we can again use (c) to write $$f(\frac{x}{x+1})=f(1)-f(\frac{1}{x+1})=f(1)-\frac{f(x)+1}{(x+1)^2}.$$So we have $$f(1)-\frac{f(x)+1}{(x+1)^2}=\frac{f(x)+f(1)x^2}{(x+1)^2}.$$In which the coefficient of $f(x)$ is non-zero. Condition (1) was not necessary after all.