$\begin{array}{l}\text { 3. Let } \frac{x}{x^{2}+x+1}=a \text {, where } a \neq 0 \text {. Then } \frac{x^{2}}{x^{4}+x^{2}+1} \\ =\end{array}$
$$
\begin{array}{l}
\text { 3. } \because a \neq 0 \text {, hence } x \neq 0 \text {, then we have } \frac{1}{a}=\frac{x^{2}+x+1}{x}=x+\frac{1}{x} \\
+1, \therefore x+\frac{1}{x}=\frac{1}{a}-1 \text {. Thus, } \frac{x^{4}+x^{2}+1}{x^{2}}=x^{2}+1+\frac{1}{x^{2}} \\
=\left(x+\frac{1}{x}\right)^{2}-1=\left(\frac{1}{a}-1\right)^{2}-1=\frac{1-2 a+a^{2}}{a^{2}}-1= \\
\frac{1-2 a}{a^{2}}, \therefore \frac{x^{2}}{x^{4}+x^{2}+1}=\frac{a^{2}}{1-2 a} .
\end{array}
$$
3. $\because a \neq 0$, hence $x \neq 0$, then we have $\frac{1}{a}=\frac{x^{2}+x+1}{x}=x+\frac{1}{x}$