6.150. $\frac{u^{2}}{2-u^{2}}+\frac{u}{2-u}=2$.
Solution.

Domain of definition: $\left\{\begin{array}{l}u \neq 2, \\ u \neq \pm \sqrt{2} .\end{array}\right.$

From the condition we have $\frac{u^{2}(2-u)+u\left(2-u^{2}\right)-2\left(2-u^{2}\right)(2-u)}{\left(2-u^{2}\right)(2-u)}=0 \Leftrightarrow$ $\Leftrightarrow \frac{2 u^{3}-3 u^{2}-3 u+4}{\left(2-u^{2}\right)(2-u)}=0 \Leftrightarrow 2 u^{3}-3 u^{2}-3 u+4=0$.

We can represent the equation as $\left(2 u^{3}-2 u^{2}\right)-\left(u^{2}-u\right)-(4 u-4)=0$, $2 u^{2}(u-1)-u(u-1)-4(u-1)=0,(u-1)\left(2 u^{2}-u-4\right)=0$, then $u-1=0$, or $2 u^{2}-u-4=0$, from which $u_{1}=1 ; u_{2,3}=\frac{1 \pm \sqrt{33}}{4}$.

Answer: $u_{1}=1 ; u_{2,3}=\frac{1 \pm \sqrt{33}}{4}$.