HEEEELP

We can actually find $f^{-1}(x)$ explicitly.

$\begin{array}{rcl} f(x) &=& \dfrac{x - 3}{x - 4}\\ f\left(\color{red}f^{-1}(x)\right) &=& \dfrac{\color{red}f^{-1}(x)\color{black} - 3}{\color{red}f^{-1}(x)\color{black} - 4}\\ x &=& \dfrac{f^{-1}(x) - 3}{f^{-1}(x) - 4}\\ x(f^{-1}(x) - 4) &=& f^{-1}(x) - 3\\ xf^{-1}(x) - 4x &=& f^{-1}(x) - 3\\ (x - 1)f^{-1}(x) &=& 4x - 3\\ f^{-1}(x) &=& \dfrac{4x - 3}{x - 1} \end{array}$

Therefore, $f^{-1}(x)$ is undefined when the denominator is 0, i.e., x - 1 = 0. Solving gives x = 1 as the answer.