Show a function is continuous on $\mathbb{R}$ by any method Let $f(x)=\frac{Kx}{K^2+x^2}$ where K is some constant, show this is continuous on $\mathbb{R}$.
Here are my scratch work in looking for a delta. let $x,y\in \mathbb{R} $ WTS $|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}|<\epsilon,\forall \epsilon>0$ whenever $|x-y|<\delta$. So, $|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}|=|K||\frac{x(K^2+y^2)-y(K^2+x^2)}{(K^2+y^2)(K^2+x^2)}|\leq |\frac{xK^2-yK^2+xy^2-yx^2}{(K^2+y^2)(K^2+x^2)}|\leq|\frac{xK^2-yK^2+xy^2}{(K^2+y^2)(K^2+x^2)}|$ I'm a bit stuck here on how to relate this inequality to $|x-y|$. Can someone help out? Thanks!

Though you agreed to use a more appropriate general argument (see comments), let us finish your $\epsilon-\delta$-proof (assuming $K\ne0$).
You were stuck on how to relate the inequality
$$\left|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}\right|\le|K|\left|\frac{xK^2-yK^2+xy^2-yx^2}{(K^2+y^2)(K^2+x^2)}\right|$$
to $|x−y|.$ Just factorize it in the numerator:
$xK^2-yK^2+xy^2-yx^2=(x-y)(K^2-xy)$
hence for $|x-y|<\delta$,
$$\left|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}\right|\le\delta|K|M\quad\text{where}\quad M:=\frac{K^2+|xy|}{(K^2+y^2)(K^2+x^2)}$$
and you just need to bound $M$.
If you only want to prove the continuity at any fixed point $x$, you can end up like this: wlog $\delta<1$, so that
$$M\le\frac{K^2+|x|(|x|+1)}{K^4}.$$
But you can even prove that $f$ is uniformly continuous:
$$M\le\frac{K^2}{K^4}+\frac{|xy|}{K^2(x^2+y^2)}\le\frac1{K^2}+\frac1{2K^2}.$$
