Solution: Since $x+y+z-1=0$. Introduce a parameter $t>0$.
Then
$$\begin{array}{l} \frac{1}{x}+\frac{4}{y}+\frac{9}{z} \\ =\frac{1}{x}+\frac{4}{y}+\frac{9}{z}+t(x+y+z-1) \\ =\frac{1}{x}+t x+\frac{4}{y}+t y+\frac{9}{z}+t z-t \\ \therefore \frac{1}{x} \cdot t+2 \sqrt{\frac{4}{y}} \cdot t y+2 \sqrt{\frac{9}{z}} \cdot t z-t \\ 12 \sqrt{t}-t=36-(\sqrt{t}-6)^{2} \text {. } \\ \end{array}$$

By the arbitrariness of $t$, we know $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}=36$.
Also, when $t x=\frac{1}{x}, t y=\frac{4}{y}, t z=\frac{9}{z}, t=36$, $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}=36$,
i.e., the minimum value of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$ is 36.