How do i solve this quadratic equation by factoring ?

I'd suggest you get rid of the fraction first.  Mult. every term by 3.  You will get

3y^2 + 10y + 3 = 0

From inspection, with coefficient a=3 and coeff. c = 3, the binomial factors could possibly begin with y or 3y:  for example, y+1; also, the binom. factors may end in +1.    Let's try the possible binomial factor 3y + 1.

Note that 10y separates into 9y+1y.

Then 3y^2 +    10y     +     3  = 0 becomes
         3y^2   + 9y + 1y  +  3  =  0

Let's apply factoring by grouping:

        3y^2 + 9y + 1y + 3 = 0
        3y*(y + 3) + 1(y + 3)    so y+3 is indeed a common factor.

Factoring y+3 out, we get (y+3)(3y + 1), which prove to be the correct set of factors.  Multiply these together to ensure that the product is indeed y^2 + (10/3)y + 1.

X=3 and y=4 is the answer