Someone showed me such an algorithm. I'm going to explain his method with an example. Let's say we're looking for $$37^{-1}\in \mathbb Z/63\mathbb Z$$ First, we make the successive divisions of Euclid's algorithm and present the results in the form of the following table: \begin{matrix} \textbf{63} & \textbf{37} &\color{green}{26} &11 &4 &3 &\color{orange}1 \\ & \color{red}1 & 1 & 2 & 2 & 1 & \end{matrix} For example, $63=\color{red}1\times 37+\color{green}{26}, 37=1\times26+11...$ until $\color{orange}1=\gcd(63,37)$
Then, You start a third line where you write $1$ and $0$ And you write $\color{red}{\text{the opposites}}$ of the second line : $$\begin{matrix} 63 & 37 &26 &11 &4 &3 &1 \\ & \color{red}-1 &\color{red} -1 &\color{red} -2 &\color{red} -2 &\color{red}- 1 & \\ & & & & & & 1&0 \end{matrix}$$Then you complete the third line with a process that is a bit difficult to get used to at first but you get used to quite quickly$$\begin{matrix} 63 & 37 &26 &11 &4 &3 &1 \\ & -1 & -1 & -2 & -2 &\textbf{-1} & \\ & & & & &\color{red} {\textbf{-1}} & \textbf{1}&\textbf{0} \end{matrix}$$ You calculate $\textbf{1}\times \textbf{(-1)}+\textbf{0}$; you obtain $\color{red} {\textbf{-1}}$. Then $\textbf{-1}\times \textbf{(-2)}+\textbf{1}=\color{red}{\textbf{3}}$ and so on... You finally obtain$$\begin{matrix} 63 & 37 &26 &11 &4 &3 &1 \\ & -1 & -1 & -2 & \textbf{-2} &-1& \\ & \color{green}{-17}& 10& -7& \color{red}3& \textbf{-1}& \textbf{1}&0 \end{matrix} $$And the final result $$\boxed{37^{-1}=\color{green}{-17}\text{ in }\mathbb Z/63\mathbb Z}$$
I have used this algorithm a lot by hand, I have programmed it on a computer but I have no formal proof of its validity. Maybe you'll be interested in looking for one, after you've figured out the algorith.
My attempts have been to start with classical calculus with matrices :$$\begin{bmatrix}1 & -1 \\0 & 1\end{bmatrix}\begin{bmatrix}63 \\37\end{bmatrix}=\begin{bmatrix}26 \\37\end{bmatrix},\begin{bmatrix}1 & 0 \\-1 & 1\end{bmatrix}\begin{bmatrix}26 \\37\end{bmatrix}=\begin{bmatrix}26 \\11\end{bmatrix}$$ Hence $$\begin{bmatrix}1 & -1 \\-1 & 2\end{bmatrix}\begin{bmatrix}63 \\37\end{bmatrix}=\begin{bmatrix}26 \\11\end{bmatrix}...$$But I'm lost in all these tables and numbers. Perhaps the example is poorly chosen and we can build examples where the link will be more obvious.
