jmc

From the given equation, $$2013x[f(x-1)+f(x)+f(x+1)]=[f(x){]}^2$$for all $x\ne 0.$

Let $d$ be the degree of $f(x).$ Then the degree of $2013x[f(x-1)+f(x)+f(x+1)]$ is $d+1,$ and the degree of $[f(x){]}^2$ is $2d.$ Hence, $2d=d+1,$ so $d=1.$

Accordingly, let $f(x)=ax+b.$ Then the equation $2013x[f(x-1)+f(x)+f(x+1)]=[f(x){]}^2$ becomes $$2013x(3ax+3b)=(ax+b{)}^2.$$Since $f(x)=ax+b,$ we can write this as $[f(x){]}^2=6039xf(x),$ so $$f(x)(f(x)-6039x)=0.$$Thus, $f(x)=0$ or $f(x)=6039x.$ Since $f(x)$ is non-constant, $f(x)=6039x.$ Thus, $f(1)=\boxed{6039}.$ We can check that $f(x)=6039x$ satisfies the given equation.