It is hard to say what you might have done wrong when you do not show what you did!
While it is a good idea to calculate values until you can guess a pattern, you are not done until you have proved that pattern. At the very least, you should have checked your "guess". y_3= \sqrt{\frac{2+3}{2+ 1}}\sqrt{6}y_0= \sqrt{\frac{5(6)}{3}}y_0= \sqrt{10}y_0. 10 is NOT 3n= 3(3)= 9. Certainly, you should have realized that "y_n= \sqrt{3n}y_0" is not true when n= 0!
Yes, y_1= \sqrt{3}y_0 and y_2= \sqrt{6}y_0. But then y_3= \sqrt{10}y_0, as I showed, and then y_4= \sqrt{(3+3)/(3+ 1)}\sqrt{10}y_0= \sqrt{15}y_0, etc.6,
Is that enough to make a better guess at a pattern? I notice that it is always the square root of an integer times y_0. Further, those integers are 3, 6, 10, 15... For me, at least, that is enough to notice that the 'first differences' are 6-3= 3, 10- 6= 4, 15- 10= 5 and then the 'second differences' are 4- 3= 1 and 5- 4= 1- both 1. IF that continued to be true, that the 'second difference' is a constant, that would mean that the integers are a quadratic function of n, an^2+ bn+ cand we can use those values to determine a, b, and c.
I did that and got that a= 1/2, b= 3/2, and c= 1 so that the integers are (1/2)(n^2+ 3n+ 2)= (1/2)(n+ 2)(n+ 1), exactly what your textbook says.
You can prove that is true using "proof by induction". When n= 0, y_0= \sqrt{\frac{1}{2}(0+2)(0+ 1)}y_0= \sqrt{1}y_0= y_0 is correct. Assume that, for some k, y_k= \sqrt{\frac{1}{2}(k+2)(k+ 1)}y_0. Then, by the difference equation, y_{k+1}= \sqrt{(k+3)/(k+1)}y_k= \sqrt{\frac{1}{2}((k+3)/(k+1))(k+2)(k+1)}y_0= \sqrt{\frac{1}{2}(k+3)(k+2)}y_0= \sqrt{\frac{1}{2}((k+1)+ 2)((k+1)+ 1)}y_0.
