how to solve this linear system of three equations using Cramer's rule?

The determinant of the matrix representing $x$ is, expanding down the middle column (which you definitely can do, no row operations needed!):

$$\det\begin{pmatrix}-6&2&3\\2&0&1\\-5&1&-1\end{pmatrix}=(-2)\cdot\det\begin{pmatrix}2&1\\-5&-1\end{pmatrix}+0+(-1)\cdot\det\begin{pmatrix}-6&3\\2&1\end{pmatrix}$$

Which is: $$(-2)(-2-(-5))+(-1)(-6-6)=-2\cdot3+12=6$$Cramer tells that $x$ is then this determinant divided by the original one, namely $6/6=1$. When you’re saying: “this matrix divided by the determinant of the original matrix” this is wrong. That would imply $x$ is a matrix divided by a scalar, i.e. another matrix. You just take the determinants of both, and find their ratios.

The determinant of$$A_x=\begin{bmatrix}-6&2&3\\2&0&1\\-5&1&-1\end{bmatrix}$$is $6$, which is equal to $\det(A)$. Therefore, $\frac{\det(A_x)}{\det(A)}=1$, which is what you should have got.