How can we verify this limit via ϵ−δ method? I have to calculate the limit...

We have to show that
${\displaystyle \mathrm{\forall }ϵ>0:\mathrm{\exists }\delta \in (0,\frac{\pi }{2}]:\mathrm{\forall }x\in R\left\{0\right\}}$
${\displaystyle :(\left|x\right|<\delta \Rightarrow \left|f\left(x\right)\right|<ϵ)}$
where ${\displaystyle f\left(x\right)=\frac{{(1-\cos x)}^2}{6{\sin }^{2}x}=\frac{{(1-\cos x)}^2}{6(1-{\cos }^{2}x)}}$
Let ${\displaystyle t=\cos x}$. Then, because ${\displaystyle t\ne 1\text{whenever}x\ne 0\text{and}\left|x\right|<\frac{\pi }{2}}$,
${\displaystyle f\left(x\right)=\frac{{(1-t)}^2}{6(1-t^2)}=\frac{(1-t)}{6(1+t)}=\frac{(2-1-t)}{6(1+t)}}$
${\displaystyle =\frac{1}{3(1+t)}-\frac{1}{6}}$
f(x) strictly decreases w.r.t.t.
Find t that makes ${\displaystyle f\left(x\right)=ϵ}$. After all, ${\displaystyle t=\frac{2}{(6ϵ+1)}-1}$. Since ${\displaystyle t=\cos x}$, choosing
${\displaystyle \delta =\arccos (\frac{2}{6min\{ϵ,\frac{1}{6}\}+1}-1)}$
proves the claim. I think you can show that ${\displaystyle \delta }$ is an increasing function of ${\displaystyle ϵ}$.

You can also multiply and divide by ${\displaystyle 1+\cos x}$:
${\displaystyle \underset{x\to 0}{lim}\frac{1-\cos x}{3{\sin }^{2}x}=\underset{x\to 0}{lim}\frac{1-{\cos }^{2}x}{3{\sin }^{2}x(1+\cos x)}=}$
${\displaystyle \underset{x\to 0}{lim}\frac{{\sin }^{2}x}{3{\sin }^{2}x(1+\cos x)}=\frac{1}{3}\underset{x\to 0}{lim}\frac{1}{1+\cos x}=\frac{1}{6}}$
If you need to use the definition, you can easily see that
${\displaystyle |\frac{1-\cos x}{3{\sin }^{2}x}-\frac{1}{6}|=\frac{1}{3}|\frac{1}{1+\cos x}-\frac{1}{2}|\to 0(x\to 0)}$
Since there is no indetermination in this limit, you can use Heine's definition in a very straightforward way (instead of Cauchy's). Or, you can go on to obtain
${\displaystyle |\frac{1}{1+\cos x}-\frac{1}{2}|=\frac{1}{2}\left|\frac{1-\cos x}{1+\cos x}\right|\le \frac{1}{2}|1-\cos x|\le \frac{1}{2}\left|x\right|}$