Plug in $k=-1$. That gives you a basis for $\mathbb{R}^3$, as discussed in your previous question, namely $\left(\begin{array}{r}1\\2\\-1\end{array}\right),\qquad\left(\begin{array}{r}2\\0\\-1\end{array}\right),\qquad\left(\begin{array}{r}-3\\1\\5\end{array}\right).$ That's the "basis in the domain".
You know what the image of the basis vectors under $f_{-1}$ is: you are told what they are.
So you have a basis $\beta$, and the value of the linear transformation at the vectors of $\beta$. How do you find the matrix of $f_{-1}$ from $\mathbb{R}^3$ with basis $\beta$ to $\mathbb{R}^3$ with the standard basis? The first column is the image of the first vector of $\beta$ written in terms of the standard basis. The second column of the matrix is...
Edit. I see now that the question asks for the matrix of $f_{-1}$ relative to the standard bases, which means you need to find $f_{-1}(e_1)$, $f_{-1}(e_2)$, and $f_{-1}(e_3)$. How to do that?
Well, since $(1,2,-1)$, $(2,0,-1)$, and A$(-3,1,5)$ are a basis, we can write $e_1$, $e_2$, and $e_3$ as linear combinations of them. For example, $e_1 = -\frac{1}{15}\left(\begin{array}{r}1\\2\\-1\end{array}\right) + \frac{11}{15}\left(\begin{array}{r}2\\0\\-1\end{array}\right) + \frac{2}{15}\left(\begin{array}{r}-3\\1\\5\end{array}\right)$ so that means that $f_{-1}(e_1) = -\frac{1}{15}f_{-1}\left(\begin{array}{r}1\\2\\-1\end{array}\right) + \frac{11}{15}f_{-1}\left(\begin{array}{r}2\\0\\-1\end{array}\right)+ \frac{2}{15}f_{-1}\left(\begin{array}{r}-3\\1\\5\end{array}\right).$ Continue this way to get the matrix.