The first few convergents of $[3;\overline{1,6}]$ are $3,4,\frac{27}7,\frac{31}8,\frac{213}{55}$, and $\frac{244}{63}$; $\frac{213}{55}=3.8\overline{72}$, and $\frac{244}{63}$ is a little over $3.873$, so $[3;\overline{1,6}]$ is clearly not $\sqrt{3^2-2\cdot3}=\sqrt3$.
In fact, $n^2-2n=(n-1)^2-1$, so the integer part and first convergent of $\sqrt{n^2-2n}$ will be $n-2$. But $\sqrt{n^2-2n}$ isn’t $[n-2;\overline{1,2n}]$, either, since, as you can see here, $\sqrt3=[1;\overline{1,2}]$.
$\sqrt8=[2;\overline{1,4}]$, so your identity should probably be $\sqrt{n^2-2n}=[n-2;\overline{1,2(n-2)}]$ for $n\ge 3$.
Now let $m=n-2$ and consider the continued fraction $x=[m;\overline{1,2m}]$.
$\begin{align*} x+m&=[2m;\overline{1,2m}]\\ &=2m+\frac1{1+\frac1{[2m;\overline{1,2m}]}}\\ &=2m+\frac1{1+\frac1{x+m}}\\ &=2m+\frac{x+m}{x+m+1}\\ &=\frac{(2m+1)x+2m^2+3m}{x+m+1}\;, \end{align*}$
and you can finish it off by solving for $x$ in terms of $m$ (and then in terms of $n$).
If the identity was supposed to have $\sqrt{n^2+2n}$ on the lefthand side, the same basic approach will work, though the details will obviously be different, and you won’t need $m$.
Added: I see that you have edited the problem statement to make the lefthand side $\sqrt{n^2+2n}$; I’ll leave my solution to the original version as an extended hint for the corrected version.