Some of the key things to consider here are that the domain of the logarithmic function must be greater than zero, and we cannot divide by zero.
ln(3) is a constant - it just moves the locus ("graph") of the equation, so let's focus on the rest for now.
x may not equal -3, or 3+x would equal zero, and we can't have that. Neither may x equal -5, or the numerator 5+x would equal zero, and we would have to take the logarithm of zero. Finally, x cannot be between -3 and -5, or the quotient (5+x)/(3+x) will be negative, and we would have to take the logarithm of a negative number.
So, the domain is: x<-5 U -3<x (Read that "x is less than negative five union negative 3 is less than x.")
The range is a little more problematic.
Since we are subtracting ln(3) from the result, the locus of the equation will be asymptotic to that value, which is about equal to -1.099. In other words, as the locus of the graph goes off to infinity on the negative x side, the values of y will be negative, but will increase, getting closer and closer to -1.099, but never actually getting there. Similarly, on the right side, as the x's are positive and get bigger, the values of y will approach -1.099 from above, getting smaller and smaller, but never quite reaching this number.
Otherwise, as the graph approaches -5 from the left, the domain will get arbitrarily small (in other words, big negative numbers), and, similarly, as the graph approaches -3 from the right, the domain will get arbitrarily big.
So, the range is y may not equal ln(3).