Answer
$y=-2x\pm3$
Work Step by Step
Step 1. Identify the slope of the line $2x+y=0$ as $m_1=-2$
Step 2. Identify the slope of the normal line $m_2=m_1=-2$
Step 3. Identify the slope of the tangent line to the curve $m_3=-\frac{1}{m_2}=\frac{1}{2}$
Step 4. Differentiate both sides of the curve $xy+2x-y=0$ with respect to $x$: $y+xy'+2-y'=0$, which gives
$y'=\frac{y+2}{1-x}$
Step 5. Letting $y'=m_3$, we have $\frac{y+2}{1-x}=\frac{1}{2}$, which gives $2y+4=1-x$ or $x=-2y-3$
Step 6. Use the above relation with the equation of the curve: $(-2y-3)y+2(-2y-3)-y=0$. We get $-2y^2-3y-4y-6-y=0$ or $2y^2+8y+6=0$ or $y^2+4y+3=0$, which gives $y=-1, -3$
Step 7. For $y=-1$, $x=-1$, for $y=-3$, $x=3$, thus the two points are $(-1,-1)$ and $(3,-3)$
Step 8. The line equations for the normal are: $y+1=-2(x+1)$ or $y=-2x-3$ and $y+3=-2(x-3)$ or $y=-2x+3$
