Điều kiện: \left\{\begin{array}{l}5x+6>0\\x\ne0\end{array}\right.
Ta có:
\log_3^2(5x+6)^2-\log_3(5x+6)^3\log_3x^6+2\log_3^2\dfrac{1}{x^2}=0
\Leftrightarrow 4\log_3^2(5x+6)-9\log_3(5x+6)\log_3x^2+2\log_3^2x^2=0
\Leftrightarrow [4\log_3(5x+6)-\log_3x^2][\log_3(5x+6)-2\log_3x^2]=0
\Leftrightarrow \left[\begin{array}{l}4\log_3(5x+6)=\log_3x^2\\\log_3(5x+6)=2\log_3x^2\end{array}\right.
\Leftrightarrow \left[\begin{array}{l}(5x+6)^4=x^2\\5x+6=x^4\end{array}\right.
\Leftrightarrow \left[\begin{array}{l}x=-1\\x=-\dfrac{36}{25}\\x=2\end{array}\right.
Kết hợp với điều kiện ta có: x\in\{-1;2\}