Ну, решим так:
log(3x,3/x)+log(3,x)^2=1
(log(3x,3/x)+1)+log(3,x)^2=2
log(3x,9)+log(3,x)^2=2
1/log(9,3x)+log(3,x)^2=2
1/(1/2 * (1+log(3,x)))+log(3,x)^2=2
2/(1+log(3,x))+log(3,x)^2=2
Замена: log(3,x)=t
2/(1+t)+t^2=2
2+(1+t)*t^2=2*(1+t)
t^3+t^2-2t=0
t(t^2+t-2)=0
Получим три решения:
t=0 => x=3^0=1
t=1 => x=3^1=3
t=-2 => x=3^(-2)=1/9