Eigenvalue of a given operator If $u_0$ is a positive radial symmetric nontrival solution of 
$$
-\frac{1}{2}\frac{d^2u}{dx^2}+\lambda u -u^3=0
$$
Then how to show $-3\lambda$ is a eigenvalue of 
$$
Lu=-\frac{1}{2}\frac{d^2u}{dx^2}+\lambda u -3u_0^2 u
$$
and the corresponding eigenfunction is $u_0^2$ ?
What I try:
\begin{align}
Lu_0^2
&=-\frac{1}{2}\frac{d^2u_0^2}{dx^2}+\lambda u_0^2 -3u_0^4 \\
&=-(\frac{du_0}{dx})^2 -(u_0 \frac{d^2 u_0}{dx^2}) + \lambda u_0^2 -3 u_0^4  \\
&=-(\frac{du_0}{dx})^2-\frac{1}{2}(u_0 \frac{d^2 u_0}{dx^2})-2u_0^4
+ [-\frac{1}{2}(u_0 \frac{d^2 u_0}{dx^2})+ \lambda u_0^2 - u_0^4]  \\
&=-(\frac{du_0}{dx})^2-\frac{1}{2}(u_0 \frac{d^2 u_0}{dx^2})-2u_0^4
\end{align}
then I don't how to deal the $-(\frac{du_0}{dx})^2$, what should I do ?

Here is my try.  Let $v_0=\frac{du_0}{dx}$.  From
$$-\frac12\frac{d^2u_0}{dx^2}+\lambda u_0-u_0^3=0\implies \frac{d^2u_0}{dx^2}=2\lambda u_0-2u_0^4,$$
we have as you computed
$$Lu_0^2=-v_0^2-u_0\frac{d^2u_0}{dx^2}+\lambda u_0^2-3u_0^4=-v_0^2-u_0(2\lambda u_0-2u_0^4)+\lambda u_0^2-3u_0^4.$$
So $Lu_0^2=-v_0^2-\lambda u_0^2+u_0^4$.  Also from
$$-\frac12\frac{d^2u_0}{dx^2}+\lambda u_0-u_0^3=0\implies-\frac{1}{2}v_0\frac{dv_0}{dx}+\lambda u_0\frac{du_0}{dx}-u_0^3\frac{du_0}{dx}=0,$$
we get by integrating wrt $x$ that
$$-\frac{v_0^2}{4}+\frac{\lambda u_0^2}{2}-\frac{u_0^4}{4}=c$$
for some constant $c$.  So $-v_0^2+2\lambda u_0^2-u_0^4=4c$.  That is,
$$Lu_0^2=-v_0^2-\lambda u_0^2+u_0^4=(-v_0^2+2\lambda u_0^2-u_0^4)-3\lambda u_0^2=4c-3\lambda u_0^2.$$
Presumably, you have some conditions to make $c=0$ (is there anything missing?), because if $c=0$, $Lu_0^2=-3\lambda u_0^2$ as required.
