First compute the characteristic polynomial,
\begin{align*}
\charpoly{C}{x}
&=\detname{C-xI_2}&&\knowl{./knowls/definition.CP.knowl}{\text{Definition CP}}\\
&=
\begin{vmatrix}
-1-x & 2 \\
-6 & 6-x
\end{vmatrix}\\
&=(-1-x)(6-x)-(2)(-6)&&\knowl{./knowls/theorem.DMST.knowl}{\text{Theorem DMST}}\\
&=x^2-5x+6\\
&=(x-3)(x-2)
\end{align*}
So the eigenvalues of $C$ are the solutions to $\charpoly{C}{x}=0$, namely, $\lambda=2$ and $\lambda=3$. Each eigenvalue has a factor that appears just once in the characteristic polynomial, so $\algmult{C}{2}=1$ and $\algmult{C}{3}=1$.
To obtain the eigenspaces, construct the appropriate singular matrices and find expressions for the null spaces of these matrices.
\begin{align*}
\lambda&=2\\
C-(2)I_2&=
\begin{bmatrix}
-3 & 2\\
-6 & 4
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & -\frac{2}{3}\\
0 & 0
\end{bmatrix}\\
\eigenspace{C}{2}&=\nsp{C-(2)I_2}=
\spn{\set{\colvector{\frac{2}{3}\\1}}}
=
\spn{\set{\colvector{2\\3}}}
\end{align*}
\begin{align*}
\lambda&=3\\
C-(3)I_2&=
\begin{bmatrix}
-4 & 2\\
-6 & 3
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & -\frac{1}{2}\\
0 & 0
\end{bmatrix}\\
\eigenspace{C}{3}&=\nsp{C-(3)I_2}=
\spn{\set{\colvector{\frac{1}{2}\\1}}}
=
\spn{\set{\colvector{1\\2}}}
\end{align*}
Each eigenspace has a single basis vector, so the dimensions are both $1$ and the geometric multiplicities are $\geomult{A}{2}=1$ and $\geomult{A}{3}=1$.