If someone could run quickly through the theory and methods on this it would be hugely appreciated. Thank you.
Let $f: \Bbb R^2 → \Bbb R^2$ be reflection in the line $y = x$ and let $g: \Bbb R^2 → \Bbb R^2$ be clockwise rotation through $90^\circ$ around the origin.
Find the point $v = (x,y) \in\Bbb R^2$ such that $f(g(v)) = (−3,−2)$.
$f$ and $g$ are both invertible, so to find $v$ such that $f(g(v))=(-3,-2)$, you can just invert the transformations one at a time:
$$fg(v)=(-3,-2)\\ \iff g(v)=f^{-1}((-3,-2))\\ \iff v=g^{-1}f^{-1}((-3,-2)) $$
Naturally, $f^{-1}$ is just $f$ itself, since reflections are their own inverses, and $g^{-1}$ is the counterclockwise rotation by $90^\circ$.
Thus $f^{-1}((-3,-2))=(-2,-3)$ and $g^{-1}f^{-1}((-3,-2))=g^{-1}((-2,-3))=(3,-2)$.
If you are familiar with transformations via matrices (acting on the left of column vectors) then you can check that
The matrix for $f$ and $f^{-1}$ is $\begin{bmatrix}0&1\\1&0\end{bmatrix}$
and the matrix for $g$ is $\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ and for $g^{-1}$ is $\begin{bmatrix}0&-1\\1&0\end{bmatrix}$. The composition $g^{-1}f^{-1}$ via matrix multiplication gives $\begin{bmatrix}-1&0\\0&1\end{bmatrix}$.