Помогите решить неравенство

$frac{log_{1-x}(3x+1)(1-2x+ x^{2})}{log_{3x+1}(1-x)} leq -1;$
ОДЗ: $[left{begin{aligned}&1-x textgreater 0;1-x neq 1; &3x+1 textgreater 0;3x+1 neq 1; &(3x+1)(1-2x+ x^{2}) textgreater 0; end{aligned}right.]$ $[left{begin{aligned}&x textless 1;x neq 0; &x textgreater - frac{1}{3} ;xneq0; &3(x+ frac{1}{3} )(1-x)^{2} textgreater 0; end{aligned}right.]$ $[left{begin{aligned}&(- infty;0) cup(0;1); &(- frac{1}{3};0) cup(0; +infty); &(- frac{1}{3};1 ) cup(1;+ infty) textgreater 0; end{aligned}right.]$
$(- frac{1}{3};0) cup(0; 1)$
$(log_{1-x}(3x+1)+log_{1-x}(1-x)^{2}) cdotfrac{1}{log_{3x+1}(1-x)} leq -1;$
$(log_{1-x}(3x+1)+log_{1-x}(1-x)^{2})log_{1-x}(3x+1) leq -1;$
$(log_{1-x}(3x+1)+2)log_{1-x}(3x+1) leq -1;$
$log^2_{1-x}(3x+1)+2log_{1-x}(3x+1)+1 leq 0;$
$(log_{1-x}(3x+1)+1)^2 leq 0;$
Возможно только равенство нулю.
$log_{1-x}(3x+1)+1=0;$
$log_{1-x}(3x+1)=-1;$
$3x+1=(1-x)^{-1};$
$3x+1= frac{1}{1-x};$
$(3x+1)(1-x)= 1;$
$3 x^{2}-2x=0;x(3x-2)=0;$
$x=0;$ - посторонний корень.
$x= frac{2}{3} in(- frac{1}{3};0) cup(0; 1).$
Ответ: $x= frac{2}{3}.$