Is my particular solution of $ {(3xy + y^2)}\ dx + {(x^2 + xy)}\ dy = 0 $ for $x=1$ and $y=1$ correct? 
Find the particular solution to 
  $$
{(3xy +  y^2)}\ dx + {(x^2 + xy)}\ dy = 0
$$
for $x=1$ and $y=1$.

My Solution
$$
\frac{dy}{dx}=-\frac{3xy+y^2}{x^2+xy}
$$
Substituting $y=vx$ , $\frac{dy}{dx}=v+x\frac{dv}{dx}$
$$
v+x\frac{dv}{dx}=-\frac{3vx^2+x^2v^2}{x^2+vx^2}
$$
$$
v+s\frac{dv}{dx}=-\frac{3v+v^2}{1+v}
$$
$$
x\frac{dv}{dx}=-\frac{2v^2+4v}{v+1}
$$
$$
\frac{v+1}{2v^2+4v}=-\frac{dx}{x}
$$
$$
\frac{(2v+2)}{v^2+2v}=-4\frac{dx}{x}
$$
$$
\int\frac{(2v+2)}{v^2+2v}=-4\int\frac{dx}{x}
$$
$$
\log |v^2+2v| = -4\log|x|+\log C 
$$
$$
\log |v^2+2v| = \log\frac{C}{x^4}
$$
$$
|v^2+2v|=\frac{C}{x^4}
$$
$$
|y^2+2xy|=\frac{C}{x^2}
$$
Substituting $x=1$ and $y=1$, $C=3$
The answer is 
$$
|{(y^2 + 2xy)}|= \frac{3}{x^2}
$$
The answer in the book is left till here only,but what I did after this was as follows
$$
 (y^2 + 2x y) = \pm\frac{3}{x^2}
$$
And then rejecting 
$$
 (y^2 + 2x y) = -\frac{3}{x^2}
$$
As it does not satisfy $x=1$, $y=1$
So According to me the particular solution is 
$$
 (y^2 + 2x y) = \frac{3}{x^2}
$$
Whose answer is correct my or book's?
 A: Neither you nor the book is correct.  Both are partially correct, but not correct entirely.  The false step is the belief that
$$\int\,\frac{1}{t}\,\text{d}t=\ln|t|+\text{constant}\,.$$
Since $t=0$ is not in the domain of the function $t\mapsto\dfrac1t$, what is correct is saying that
$$\int\,\frac1t\,\text{d}t=\ln|t|+c(t)\,,$$
where $c:(-\infty,0)\cup(0,\infty)\to\mathbb{C}$ is a locally constant function.  That is, for some constants $c_-$ and $c_+$ which are not necessarily the same, we have
$$c(t)=\begin{cases}c_-&\text{if }t<0\,,\\c_+&\text{if }t>0\,.\end{cases}$$
Therefore, what is safe to say is that
$$|v^2+2v|=\frac{\tilde{\Gamma}(x,v)}{x^4}$$
for some locally constant function $\tilde{\Gamma}$ in two variables with domain $\Omega:=\big(\mathbb{R}\setminus\{0\} \big)\times \big(\mathbb{R}\setminus\{-2,0\}\big)$.  You can remove the absolute value and say
$$v^2+2v=\frac{\Gamma(x,v)}{x^4}$$
for some locally constant $\Gamma:\Omega\to\mathbb{C}$. 
From the initial condition $(x,y)=(1,1)$ (whence $(x,v)=1$) that you are dealing with the connected component $(0,\infty)\times (0,\infty)$ of $\Omega$.  You can take $\Gamma(x,v)$ to be a constant $C$ there, which you found out that $C=3$.  That is,
$$y^2+2xy=x^2(v^2+2v)=\frac{C}{x^2}=\frac{3}{x^2}\text{ for }x>0\,.$$
However, you cannot jump the boundary and deduce that $y^2+2xy=\dfrac{3}{x^2}$ for $x<0$ too.
