Solving $13\alpha \equiv 1 \pmod{210}$

CRT $ $ (Chinesese Remainder Theorem) $ $ yields a handy inverse reciprocity formula - which enables us to calculate $\rm\ \ x^{-1}\bmod y\ \ $ from $\rm\ \ y^{-1}\bmod x,\,$ namely

$\begin{align} \rm CRT\ \Rightarrow\rm \ x\,(x^{-1}\bmod y)\ &\rm +\ y\,(y^{-1}\bmod x)\ \equiv\ 1\!\pmod{xy}\quad when\quad (x,y)=1 \\[.5em] \Rightarrow \rm\ \ \ \ \ \ x^{-1}\bmod y\ \ &\equiv\: \rm \dfrac{1\ +\ y\, (\color{#c00}{-y^{-1}\bmod x})}x\!\pmod{\! y}\quad\ {\bf [Inverse\ Reciprocity]} \\[.5em] \Rightarrow\! \rm\ \ 13^{-1}\bmod 210\ & \equiv\ \rm \dfrac{1\ +\ 210\ \ (-210^{-1}\,mod\ 13)}{13} \\[.5em] &\equiv\rm\ \dfrac{1 + (\color{#90f}2 + 13\cdot 16)\,(-{\color{#90f}2}^{-1}\!\!\equiv \color{#0A0}{6}\,\bmod 13)}{13\ \ }\\[.5em] &\equiv\rm\ \dfrac{1 + (\color{#90f}{2})\,\color{#0A0}{6}}{13}\, +\, (16)\,\color{#0A0}{6} \,\equiv\, 97\!\pmod{\!210} \end{align}$

To remember it, view it fractionally as forcing $\dfrac{1}x\pmod{\!y}\,$ to be an exact quotient by replacing the top $\,1\,$ by a congruent integer $\,1+yk\,$ where $\,x\mid 1+yk,\,$ i.e. $\color{#c00}{\bmod x}\!:\, yk\equiv -1,\,$ so $\,\color{#c00}{k\equiv -y^{-1}}.\,$ Rather than computing $\,y^{-1},\,$ often we can "twiddle" the top by testing if small values of $\,k\,$ work.

See here and here and here for many more worked examples of this (and many other) methods to compute modular inverses and fractions.

Remark $\ $ Once one knows the method, the calculation amounts to the prior two lines, which is a bit quicker than using the EEA = extended Euclidean Algorithm (which is, essentially, equivalent). For example, compared to the back-substitution EEA in DonAntonio's answer we see that the above uses smaller numbers, because the form of the inversion formula allows us to cancel $13$ before performing any multiplications. Generally the arithmetic will be a bit simpler due to this. For completeness, here is the inversion algorithm and a complete proof (without using CRT).

$\begin{eqnarray} {\bf Lemma}\ \ &&\rm y &=&\rm \ \ \,r \, +\, q x &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\rm e.g.\ &\ \ 210\ \,= \ \:2 +\! 16\cdot 13 &&\rm\ \ (1)\ \ find\ \ r \ =\,\ y\ \ \,mod \ x\\ &&\rm 1 &=\,&\rm r'r\, +\, p x & &\ \ 1 =\! (-6)2 \,+\, \color{#C00}1\cdot 13 &&\rm\ \ (2)\ \ find\ \ r' = r^{-1}\ mod\ x \\ \Rightarrow &&\rm 1 &\equiv&\rm x\,(\color{#0A0}{p - q r'})\ \ \ (mod\ y)& &\ \ 1 \equiv 13(\color{#C00}1-16(-6)) &&\rm\ \ (3)\ \ find\,\ x' = x^{-1}\, mod\ y\\ {\bf Proof\quad\ } && &=&\rm 1\!-\!rr'\!\!-\!xqr'\! & &\ \ \phantom{1} \equiv 13\cdot 97&&\rm\ \ \phantom{(3)\ find }x^{-1}\! =\, determinant\\ && &=&\rm 1\!-\!(r\!+\!qx)r'\! = 1\!-\!yr'\!\equiv 1\!\!\!\!\!\!\! && &&\rm\ \ \phantom{(3)\ \ find\,\ x' =} \ (see\ below) \end{eqnarray}$

The algorithm is very easy to remember since the $\rm\color{#0A0}{formula}$ for $\rm\:x^{-1}\:$ is just the determinant of the system of linear equations! Indeed, the Lemma has a more insightful proof by Cramer's Rule.

$\begin{eqnarray} \begin{array}{l}{\bf Lemma}\\ \phantom{.}\end{array}\quad\ \begin{array}{l} \rm y = \ \ \, r + q\,x\\ \rm 1 = r'r + p\,x\end{array}\ \Rightarrow\ \left[\begin{array}{c}\rm y\\ \rm 1\end{array}\right]\! &=& \left[\begin{array}{cc}\rm 1 &\rm q\\ \rm r' &\rm p\end{array}\right] \left[\begin{array}{c}\rm r\\ \rm x\end{array}\right]\ \Rightarrow\,\ \rm x^{-1}\! = det = p\!-\!qr'\ \ (mod\ y) \\ \\ \begin{array}{l}\rm {\bf Proof}\quad\ \ By\ Cramer\\ \phantom{.}\end{array}\ \ \ \ \ \ \left[\begin{array}{cc}\rm 1 &\rm y\\ \rm r' &\rm 1\end{array}\right]\!&=&\left[\begin{array}{cc}\rm 1 &\rm q\\ \rm r' &\rm p\end{array}\right] \left[\begin{array}{cc}\rm 1 &\rm r\\ \rm 0 &\rm x\end{array}\right]\\ \\ \rm Taking\ \ det\ \Rightarrow\ 1-yr' &=&\rm\ (p-qr')\ x \quad \Rightarrow\quad 1 \,\equiv\, (p-qr')\,x\ \ \ (mod\ y)\end{eqnarray}$

$$210=13\cdot 16+2\\13=6\cdot2 +1\\2=2\cdot 1$$

So going backwards:

$$1=13-6\cdot 2=13-6(210-13\cdot 16)=97\cdot 13+(-6)\cdot210\Longrightarrow$$

$$\Longrightarrow 13^{-1}=97\pmod{210}$$

You correctly determined $13$ as the only invertible number in the given range.

Hint: The result of the extended Euclidean algorithm should have the form $13a + 210b = 1$.

EDIT: Looking at DonAntonio's answer, I think you only did the first step of the Euclidean algorithm. You need to repeat the division with remainder until eventually you get the remainder $0$ (or if you overlook the consequences, when you get $1$).