The Sylvester formula, as you write, is that if $A$ is diagonalizable, and $f(t)$ is a polynomial (in fact, any analytic function), then $f(A) = \sum_{i=1}^n f(\lambda_i)A_i$ where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$ (hence, $n$ is the size of the matrix), and $A_i$ is the Frobenius covariant of $A$, $A_i = \prod_{\stackrel{j=1}{j\neq i}}^n \frac{1}{\lambda_i-\lambda_j}(A-\lambda_jI).$
Here, your matrix is $2\times 2$, so you will have $f(A) = f(\lambda_1)A_1 + f(\lambda_2)A_2,$ where $A_1 = \frac{1}{\lambda_1-\lambda_2}(A-\lambda_2I),\qquad A_2 = \frac{1}{\lambda_2-\lambda_1}(A-\lambda_1I).$
So the first step is to find the two eigenvalues; then compute the two Frobenius covariants; then apply the formula to $f(t)=t^{100}$.
You can compute the characteristic polynomial directly (it's $(2-t)^2 - 1 = t^2-4t+3$) and find the eigenvalues; or any of the standard tricks: the determinant of $A$ is $3$, the trace is $4$, so you want to numbers that add up to $4$ and whose product is $3$: $\lambda_1=1$ and $\lambda_2 = 3$ will do.
So then $\begin{align*} A_1 &= \frac{1}{1-3}(A-3I) = -\frac{1}{2}\left(\begin{array}{rr} -1&1\\ 1& -1 \end{array}\right)\\ A_2 &= \frac{1}{3-1}(A-I) = \frac{1}{2}\left(\begin{array}{cc} 1&1\\ 1& 1 \end{array}\right). \end{align*}$ So the Sylvester formula tells you that $A^{100} = f(A) = \left(-\frac{f(1)}{2}\left(\begin{array}{rr}-1&1\\1&-1\end{array}\right)\right) + \left(\frac{f(3)}{2}\left(\begin{array}{cc}1&1\\1&1 \end{array}\right)\right).$
Alternatively, $A$ is diagonalizable, with eigenvectors $(1,-1)$ (corresponding to $\lambda_1=1$) and $(1,1)$ (corresponding to $\lambda_2=3$). That means that $ \left(\begin{array}{rr} \frac{1}{2} & -\frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{array}\right) A \left(\begin{array}{rr} 1 & 1\\ -1 & 1 \end{array}\right) = \left(\begin{array}{cc}1 & 0\\0 & 3 \end{array}\right).$ Since $(PAP^{-1})^n = PA^nP^{-1}$, then $\begin{align*} \left(\begin{array}{rr} \frac{1}{2} & -\frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{array}\right) A^{100} \left(\begin{array}{rr} 1 & 1\\ -1 & 1 \end{array}\right) &= \left(\left(\begin{array}{rr} \frac{1}{2} & -\frac{1}{2}\\ \frac{1}{2} & \frac{1}{2} \end{array}\right) A \left(\begin{array}{rr} 1 & 1\\ -1 & 1 \end{array}\right) = \left(\begin{array}{cc}1 & 0\\0 & 3 \end{array}\right)\right)^{100} \\ &= \left(\begin{array}{cc} 1& 0\\ 0 & 3 \end{array}\right)^{100}. \end{align*}$ Computing the last matrix is easy, and then we just need to multiply by $\left(\begin{array}{rr}1 & 1\\-1 & 1 \end{array}\right)$ on the left and $\left(\begin{array}{rr} \frac{1}{2} & -\frac{1}{2}\\ \frac{1}{2} & \frac{1}{2}\end{array}\right)$ on the right to get the value of $A^{100}$. (But this does not use Sylvester's formula).