Riccati Differential Equation Help Show that $y(x)=\frac{1}{x}$ is a solution of the Ricatti differential equation:
$$y'+\frac{1-3xy}{x^2}+3y^2=0$$
Hence using a suitable transformation of the dependant variable, solve the differential equation when $y(1)=3$.
Any help appreciated.

here is another way to solve this problem. we will make a change of variable $$v = \frac1{xy}, y =\frac1{xv}, y' = -\frac1{x^2v}-\frac{v'}{xv^2}\tag 1 $$
subbing $(1)$ in $$y'+\frac{1-3xy}{x^2}+3y^2=0, $$ we find that 
$$-\frac1{x^2v}-\frac{v'}{xv^2} +\frac{1-3/v}{x^2}+\frac3{x^2v^2} = 0$$ multiplying by $x^2v^2$ gets us $$ -v-xv'+v^2-3v+3=0\to xv' = (v-1)(v-3)$$
leading to $$\int_{1/3}^v \frac{dv}{(v-1)(v-3)}= \frac12\int_{1/3}^v \frac{dv}{v-3} - \frac{dv}{v-1} = \int_1^x\frac{dx}x = \ln (x) $$
that is $$\ln\left(\frac{3-v}{4(1-v)}\right) = 2\ln x \to \frac{3-v}{1-v} =4x^2\to v = \frac{3-4x^2}{1-4x^2}$$ and finally $$y = \frac{1-4x^2}{x(3-4x^2)}.$$
