Example 11 Suppose $1995 x^{3}=1996 y^{3}=1997 z^{3}$, $x y z>0$, and $\sqrt[3]{1995 x^{2}+1996 y^{2}+1997 z^{2}}=$ $\sqrt[3]{1995}+\sqrt[3]{1996}+\sqrt[3]{1997}$. Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=$
(1996, National Junior High School Mathematics League)
Given that there are three independent equations (equations) and exactly three letters, it should be possible to solve for $x, y, z$, but in practice, it is quite difficult. Inspired by the "chain equality" type problem from the previous question, we can introduce the letter $k$ - try.

Assume $1995 x^{3}=1996 y^{3}=1997 z^{3}=k$, then
$$
x=\sqrt[3]{\frac{k}{1995}}, y=\sqrt[3]{\frac{k}{1996}}, z=\sqrt[3]{\frac{k}{1997}} .
$$

From the given conditions, $x, y, z, k$ have the same sign, and since $x y z>0$, we know $x, y, z, k>0$. Substituting into the last equation given, we get
$$
\sqrt[3]{\frac{k}{x}+\frac{k}{y}+\frac{k}{z}}=\sqrt[3]{\frac{k}{x^{3}}}+\sqrt[3]{\frac{k}{y^{3}}}+\sqrt[3]{\frac{k}{z^{3}}} .
$$

Rearranging, we get
$$
\sqrt[3]{k}\left(\sqrt[3]{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}-\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\right)=0 \text{. }
$$

Since $k>0$, we know $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ is a root of $\sqrt[3]{u}-u=0$.

Therefore, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$. (The other two roots $0, -1$ do not meet the conditions)