This is a separable equation. We can rewrite the equation as it follows:

#x\ y' -y = 3xy#. Separating the #x#'s and #y#'s, we have that

#xy' = y+3xy#

#xy' = (1+3x)y#

#{y'}/y = 1/x +3#.

Now that variables are separated, we can integrate both sides with respect to #x#:

#\int {y'}/y\ dx = \int ( 1/x +3)\ dx#

Considering that #y'\ dx = dy# and that the integral of a sum is the sum of the integrals, we have

#\int {dy}/y = \int 1/x\ dx + \int 3\ dx#

Solving the ingrals, we have that

#\log(y) = \log(x) + 3x+c#

Solving for #y#:

#y(x) = e^{\log(x) + 3x+c} = e^{\log(x)}\cdot e^{3x} \cdot e^c= x e^{3x} e^c#.

To determine the value of #c#, we should use the condition #y(1)=0# (if this is what you meant with #y_1=0# in your question).

I'm a little puzzled with that one, because evaluating #y# for #x=1# we have #e^3 \cdot e^c#, and this is never zero for #c \in \mathbb{R}#. I hope someone can correct me if I said something wrong, or that you can correct the question if the mistake was there.