No zero's allowed, sir.

So, this problems has those nasty square roots, and expanding and squaring looks even more ugly. The cleaner way, is to use to identity $(a-b)(a=b)=a^2-b^2!$ (Just an exclamation mark, not factorial)

Multiplying both sides by $(\sqrt{x^2+2014}+x)(\sqrt{y^2+2014}+y)$ makes the LHS become

$(\sqrt{x^2-2014}+x)(\sqrt{y^2-2014}+y)(\sqrt{x^2+2014}+x)(\sqrt{y^2+2014}+y)=(x^2+2014-x^2)(y^2+2014-y^2)=2014^2$ .

Hang on, dividing this by the given equation then gives $(\sqrt{x^2+2014}+x)(\sqrt{y^2+2014}+y)=2014$ , a very nice equation as we have the original equation too. Now expanding and equating these two equations is much better.

So equate the two equations, and after some cancelling and simplifying we then get $y\sqrt{x^2+2014}=-x\sqrt{y^2+2014}$ , since $x,y$ are nonzero, one of them must be negative as square roots are positive. Assume that $x$ is negative and $y$ is positive, then we replace $z=-x>0$ . We then obtain $y\sqrt{z^2+2014}=z\sqrt{y^2+2014}$ and after squaring, $y^2z^2+2014y^2=y^2z^2+2014z^2\iff y=z$ since we have the assumption of $y$ and $z$ being positive. This implies $y=-x$ .

Then do we go and solve for $x$ and $y$ ? No, we just go back to the desired expression. (Although the solutions are just $y=-x$ which is sufficient) Substituting $x=-y$ gives $\frac{2014x+y}{4x+y}=\frac{2013x}{3x}=\boxed{\textbf{671}}$ as $x\neq 0$ .