\(\left(x+3\right)^2=9\left(2x-1\right)^2\)
=>\(\left\lbrack3\left(2x-1\right)\right\rbrack^2=\left(x+3\right)^2\)
=>\(\left(6x-3\right)^2=\left(x+3\right)^2\)
=>\(\left(6x-3-x-3\right)\left(6x-3+x+3\right)=0\)
=>7x(5x-6)=0
=>x(5x-6)=0
=>\(\left[\begin{array}{l}x=0\\ 5x-6=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac65\end{array}\right.\)