Using the Euclidean algorithm for $\frac{3}{8}$ we have, $ \begin{align*} 3&=8(0)+3\\ 8&=3(2)+2\\ 3&=2(1)+1\\ 2&=1(2)+0 \end{align*} $ and we take the quotients as our convergents $q_1=0$, $q_2=2$, $q_3=1$, and $q_4=2$. Sometimes it's useful to look at only the first few convergents. For hand computation I've been using the following (albit odd) method.
$ \begin{array}{cc|ccc} \,&\,&0&2&1&2\\ \hline 0&1&0&1&1&3\\ 1&0&1&2&3&8 \end{array} $
The top row lists the quotients. The 0 in row two is computed by $0(0)+1=0$. Then moving right, $2(0)+1=1$. Then, $1(1)+0=1$. Then, $2(1)+1=3$. The third row always starts with 1. Here, $0(0)+1=1$. Then, $2(1)+0=2$ and $1(2)+1=3$, finally $2(3)+2=8$.
In general, taking the top and bottom number under one of the quotients (or convergents) gives the fraction up to that point. That is, $\frac{1}{3}=0+\cfrac{1}{2+\cfrac{1}{1}}$ and using them all gives the orignal fraction $\frac{3}{8}$.
Finally, the last chunk of 4 numbers, "take the diagonal" note $1(8)-3(3)=-1$ and $3(3)-1(8)=1$. In general for a fraction one of those diagonals is the (positive) GCD.