Find solution for $|x| + |y| \frac{dy}{dx}=0$ I want to solve following differential equation 
$|x| + |y| \frac{dy}{dx}=0$ 
with initial condition $y(2)=-1$.  

@Robert Z,  since the it pass through $(2,-1)$
\begin{align}
  x - y \frac{dy}{dx}=0 
\end{align}
\begin{align}
   x dx = y dy 
 \end{align}
with the initial condition $y(2)=-1$, 
I have $y^2 = x^2 - 3 $. so 
\begin{align}
y= - \sqrt{x^2-3}
\end{align}

Hint. Start by solving the Cauchy problem in the quadrant which contains the initial point $(2,-1)$ where
$$x  -y(x) y'(x)=0.$$
Edit. The solution that you obtained
$$y(x)= - \sqrt{x^2-3}$$ 
is valid for $x \in [\sqrt{3},+\infty)$ (where $x\geq 0$ and $y\leq 0$). Now extend the solution in $[0,\sqrt{3}]$ and then in $(-\infty,0]$. Note that $|x| + |y(x)| y'(x)=0$ implies that 
$y'(x)\leq 0$, that is $y$ is decreasing.
Finally the complete solution $y:\mathbb{R}\to \mathbb{R}$ should be
$$y(x)=\begin{cases}
&- \sqrt{x^2-3}&\text{if $x\in [\sqrt{3},+\infty)$,}\\
&\sqrt{3-x^2}&\text{if $x\in [0,\sqrt{3}]$,}\\
&\sqrt{3+x^2}&\text{if $x\in (-\infty,0]$.}\\
\end{cases}$$

