Given three bases $B, C$ and $D$ and linear map $f:\Bbb R^2\to\Bbb R^2$, and $x$ from $\Bbb R^2$. We also know that $[x]_B=(x_1,x_2)^T$.
$$[f]_{B\to C}=\begin{bmatrix}2 & 3 \\3 & 1 \end{bmatrix}$$
$$[f]_{D\to C}=\begin{bmatrix}1 & 3 \\-1 & 5 \end{bmatrix}$$
Find $x$ in $D$ with respect to $x_1$ and $x_2$.
$$[f]_{D\to C}=\begin{bmatrix}1&3\\-1&5\end{bmatrix}\implies [f^{-1}]_{C\to D}=\begin{bmatrix}1&3\\-1&5\end{bmatrix}^{-1}$$
$v=[f]_{B\to C}x$ gives you the image of $x$ under $f$ with respect to basis $C$. $[f^{-1}]_{C\to D}v$ gives you the vector in $\Bbb R^2$ with respect to basis $D$ whose image under $f$ is $v$ with respect to basis $C$; in other words, the representation of $x$ in $D$.
$$[x]_D=\begin{bmatrix}1&3\\-1&5\end{bmatrix}^{-1}\begin{bmatrix}2 & 3 \\3 & 1 \end{bmatrix}[x]_B=\frac18\begin{bmatrix}1&12\\5&4\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}$$