Answer
False.
Work Step by Step
Add $6$ to both sides to get $$f(x)+6=\frac{3}{4}x$$ Multiply both sides by $\frac{4}{3}$ to get $$x=\frac{4}{3}f(x)+6(\frac{4}{3})=\frac{4}{3}f(x)+8$$ Thus, $f^{-1}(y)=\frac{4}{3}y+8$. $f^{-1}(8)$ can be found using $y=8$ to get $$f^{-1}(8)=\frac{4}{3}(8)+8=\frac{32}{3}+8=\frac{32}{3}+\frac{24}{3}=\frac{56}{3}$$ This isn't equal to $0$, so the statement is false.