First of all, recall that ln(x) tends towards -infinity for x->0+.

For x<1, we have e^(2x) < e^2, hence 1/e^(2x) > 1/e^2.

For x<1, ln(x) is negative, so multiplying both sides of the inequality by ln(x) gives ln(x)/e^(2x) < ln(x)/e^2.

And then you use my initial reminder, which says that the right hand side of this last inequality will tend to -infinity, and hence so will the left hand side.