How would you find the inverse of a complex function? Let's say you had a function $f \colon \mathbb{C} \longrightarrow \mathbb{C}$ defined by
$$ f(x+ \iota y) = (3+6x)+ \iota (2−3y). $$
How would you go about finding its inverse? This provoked my interest and curiosity, as it seems you can't simply manipulate the equation to get $x+ \iota y$ on the right hand side, and then swap $f(x+ \iota y)$ with $x+ \iota y$ and solve for $f(x+\iota y)$, as that yields a different result to what my textbook gives. So, how do you tackle a problem like this?

Note that, for any complex numbers $x+\iota y$ and $u + \iota v$, we have $f ( x+ \iota y) = u + \iota v$ if and only if $x+ \iota y = f^{-1}(u+\iota v)$, or $f^{-1}(u+\iota v) = x+\iota y$.
Suppose that $f(x +\iota y) = u + \iota v$, then we have
$$
 (3+6x)+\iota (2−3y) = u + \iota v,
$$
which implies
$$
\begin{align}
3 + 6x &= u \\
2 - 3y &= v,
\end{align}
$$
and these last two equations in turn imply
$$
\begin{align}
x &= \frac{ u-3 }{ 6 } \\
y &= \frac{ 2-v }{ 3 }.
\end{align}
$$
Therefore we have
$$
f^{-1}(u+\iota v) = x+\iota y =  \frac{ u-3 }{ 6 } + \iota \frac{ 2-v }{ 3 },
$$
that is,
$$
f^{-1}(u+\iota v) = \frac{ u-3 }{ 6 } + \iota \frac{ 2-v }{ 3 }.
$$
Hence the function $f^{-1} \colon \mathbb{C} \longrightarrow \mathbb{C}$ is defined by the formula
$$
f^{-1} (x + \iota y) = \frac{ x-3 }{ 6 } + \iota \frac{ 2-y }{ 3 }.
$$
PS:
There is one more piece of calculation / verification that is needed to be carried out.
We find that, for any complex $u + \iota v$, we have
$$
\begin{align}
\left( f \circ f^{-1} \right) ( u + \iota v) &= f \left( f^{-1} (u + \iota v) \right) \\
&= f \left( \frac{ u-3 }{ 6 } + \iota \frac{ 2-v }{ 3 } \right) \\ 
&= 3 + 6 \left( \frac{ u-3 }{ 6 } \right) + \iota \left[ 2 - 3 \left( \frac{ 2-v }{ 3 } \right) \right] \\
&= u + \iota v \\
&= i_{\mathbb{C}} ( u+\iota v),
\end{align}
$$
which implies that
$$
f \circ f^{-1}  = i_{\mathbb{C}},
$$
where $i_{\mathbb{C}}$ denotes the identity function on the set $\mathbb{C}$ of complex numbers.
Similarly, for any complex number $x + \iota y$, we have
$$
\begin{align}
\left( f^{-1} \circ f \right) ( x+ \iota y) &= f^{-1} \big( f( x+ \iota y) \big) \\
&= f^{-1} \big( (3+6x) + \iota (2-3y) \big) \\
&= \frac{ ( 3+6x ) - 3 }{ 6 } + \iota \frac{ 2- (2-3y) }{ 3 } \\
&= x + \iota y \\
&= i_{\mathbb{C}} (x + \iota y),
\end{align}
$$
which implies
$$
f^{-1} \circ f = i_{\mathbb{C}}.
$$
