Помогите решить неравенство

$\frac{log_{1-x}(3x+1)(1-2x+ x^{2})}{log_{3x+1}(1-x)} \leq -1;$
ОДЗ: $\left\{\begin{aligned}&1-x\ \textgreater \ 0;1-x \neq 1; \\&3x+1\ \textgreater \ 0;3x+1 \neq 1; \\&(3x+1)(1-2x+ x^{2})\ \textgreater \ 0; \\\end{aligned}\right.$ $\left\{\begin{aligned}&x\ \textless \ 1;x \neq 0; \\&x\ \textgreater \ - \frac{1}{3} ;x\neq0; \\&3(x+ \frac{1}{3} )(1-x)^{2}\ \textgreater \ 0; \\\end{aligned}\right.$ $\left\{\begin{aligned}&(- \infty;0) \cup(0;1); \\&(- \frac{1}{3};0) \cup(0; +\infty); \\&(- \frac{1}{3};1 ) \cup(1;+ \infty)\ \textgreater \ 0; \\\end{aligned}\right.$
$(- \frac{1}{3};0) \cup(0; 1)$
$(log_{1-x}(3x+1)+log_{1-x}(1-x)^{2}) \cdot\frac{1}{log_{3x+1}(1-x)} \leq -1;$
$(log_{1-x}(3x+1)+log_{1-x}(1-x)^{2})log_{1-x}(3x+1) \leq -1;$
$(log_{1-x}(3x+1)+2)log_{1-x}(3x+1) \leq -1;$
$log^2_{1-x}(3x+1)+2log_{1-x}(3x+1)+1 \leq 0;$
$(log_{1-x}(3x+1)+1)^2 \leq 0;$
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: $x= \frac{2}{3}.$