jmc

Note that $$n^4+n^2+1=(n^4+2n^2+1)-n^2=(n^2+1{)}^2-n^2=(n^2+n+1)(n^2-n+1).$$Decomposing into partial fractions, we find that $$\frac{n}{n^4+n^2+1}=\frac{1}{2}\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right).$$Now, note that if $f(n)=\frac{1}{n^2-n+1}$, then $f(n+1)=\frac{1}{(n+1{)}^2-(n+1)+1}=\frac{1}{n^2+n+1}$. It follows that $$\sum _{n=0}^{\infty }\frac{n}{n^4+n^2+1}=\frac{1}{2}((f(0)-f(1))+(f(1)-f(2))+(f(2)-f(3))+\cdots ).$$Since $f(n)$ tends towards 0 as $n$ gets large, this sum telescopes to $f(0)/2=\boxed{\frac{1}{2}}$.