$f(x)=\frac{2 x-3}{3 x+4}$
Let $f(x)=y=\frac{2 x-3}{3 x+4}$
On cross multiplication, we get
$$
\begin{aligned}
&3 x y+4 y=2 x-3 \\
&\Rightarrow \quad x(3 y-2)=-3-4 y \\
&\Rightarrow \quad x=\frac{-3-4 y}{3 y-2} \Rightarrow x=f^{-1}(y)=\frac{-3-4 y}{3 y-2}
\end{aligned}
$$
Put $y=-\frac{4}{3}$, we get
$$
\begin{aligned}
f^{-1}\left(-\frac{4}{3}\right) &=\frac{-3-4 \times\left(-\frac{4}{3}\right)}{3\left(-\frac{4}{3}\right)-2} \\
&=\frac{-3+\frac{16}{3}}{-4-2}=\frac{7}{3 \times(-6)}=-\frac{7}{18}
\end{aligned}
$$