10. Let $a, b$ be non-zero real numbers, $x \in \mathbf{R}$. If $\frac{\sin ^{4} x}{a^{2}}+\frac{\cos ^{4} x}{b^{2}}=\frac{1}{a^{2}+b^{2}}$,
then $\frac{\sin ^{2008} x}{a^{2006}}+\frac{\cos ^{2008} x}{b^{2006}}=$
$10 \cdot \frac{1}{\left(a^{2}+b^{2}\right)^{10 \times 13}}$
Given $\frac{\sin ^{4} x}{a^{2}}+\frac{\cos ^{4} x}{b^{2}}=\frac{1}{a^{2}+b^{2}}$.
Rewrite equation (1) as
$$
\mathrm{I}=\sin ^{4} x+\cos ^{4} x+\frac{b^{2}}{a^{2}} \sin ^{4} x+\frac{a^{2}}{b^{2}} \cos ^{4} x .
$$

And $1=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}=\sin ^{4} x+\cos ^{4} x+2 \sin ^{2} x \cdot \cos ^{2} x$.
Then $\frac{b^{2}}{a^{2}} \sin ^{4} x-2 \sin ^{2} x \cdot \cos ^{2} x+\frac{a^{2}}{b^{2}} \cos ^{4} x=0$, that is,
$$
\left(\frac{b}{a} \sin ^{2} x-\frac{a}{b} \cos ^{2} x\right)^{2}=0 \text {. }
$$

Thus, $\frac{\sin ^{4} x}{a^{4}}=\frac{\cos ^{4} x}{b^{4}}$.
Let this value be $C$. From equation (1),
$a^{2} C+b^{2} C=\frac{1}{a^{2}+b^{2}}$. Therefore, $C=\frac{1}{\left(a^{2}+b^{2}\right)^{2}}$.
And $\frac{\sin ^{2008} x}{a^{2006}}+\frac{\cos ^{2008} x}{b^{2006}}=a^{2} C^{302}+b^{2} C^{302}$
$$
=\left(a^{2}+b^{2}\right) \frac{1}{\left(a^{2}+b^{2}\right)^{604}}=\frac{1}{\left(a^{2}+b^{2}\right)^{602}} .
$$