Let $\frac{2010}{x^{3}}=\frac{2011}{y^{3}}=\frac{2012}{z^{3}}, x y z>0$,
and
$$
\begin{array}{l}
\sqrt[3]{\frac{2010}{x^{2}}+\frac{2011}{y^{2}}+\frac{2012}{z^{2}}} \\
=\sqrt[3]{2010}+\sqrt[3]{2011}+\sqrt[3]{2012} .
\end{array}
$$

Find the value of $x+y+z$.
Let $\frac{2010}{x^{3}}=\frac{2011}{y^{3}}=\frac{2012}{z^{3}}=k$, obviously $k \neq 0$.
Then $2010=x^{3} k, 2011=y^{3} k, 2012=z^{3} k$.
From the given, we have
$$
\sqrt[3]{x k+y k+z k}=\sqrt[3]{x^{3} k}+\sqrt[3]{y^{3} k}+\sqrt[3]{z^{3} k},
$$

which means $\sqrt[3]{k} \sqrt[3]{x+y+z}=\sqrt[3]{k}(x+y+z)$.
Since $k \neq 0$, we have $\sqrt[3]{x+y+z}=x+y+z$.
Given that $x>0, y>0, z>0$.
Therefore, $x+y+z=1$.