jmc

Setting $x=2,$ we get $$f(2)+f\left(-\frac{3}{5}\right)=2.$$Setting $x=-\frac{3}{5},$ we get $$f\left(-\frac{3}{5}\right)+f\left(\frac{1}{7}\right)=-\frac{3}{5}.$$Setting $x=\frac{1}{7},$ we get $$f\left(\frac{1}{7}\right)+f(2)=\frac{1}{7}.$$Adding the first and third equations, we get $$2f(2)+f\left(-\frac{3}{5}\right)+f\left(\frac{1}{7}\right)=\frac{15}{7}.$$Then $2f(2)-\frac{3}{5}=\frac{15}{7},$ which means $2f(2)=\frac{96}{35},$ so $f(2)=\boxed{\frac{48}{35}}.$