#### Answer

Besides $(1,1)$, the normal also intersects the curve at $(3,-1)$.

#### Work Step by Step

$$x^2+2xy-3y^2=0$$
1) Find the derivative of the function using implicit differentiation:
$$2x+2(y+xy')-6yy'=0$$ $$x+y+xy'-3yy'=0$$ $$3yy'-xy'=x+y$$ $$y'(3y-x)=x+y$$ $$y'=\frac{x+y}{3y-x}$$
2) The slope of the tangent to the curve at $(1,1)$ is $$y'=\frac{1+1}{3\times1-1}=\frac{2}{2}=1$$
We call the slope of the normal line at $(1,1)$ $k$. The product of $k$ and $y'$ equals $-1$, as tangent line and normal line are perpendicular:
$$k\times1=-1$$ $$k=-1$$
So the normal line to the curve at $(1,1)$ is $$y-1=-(x-1)$$ $$y-1=-x+1$$ $$y=-x+2=2-x$$
Substitute $y=2-x$ back to the original function of the curve to find the intersecting point besides $(1,1)$:
$$x^2+2x(2-x)-3(2-x)^2=0$$ $$x^2+4x-2x^2-3(4-4x+x^2)=0$$ $$-x^2+4x-12+12x-3x^2=0$$ $$-4x^2+16x-12=0$$ $$x^2-4x+3=0$$ $$x=1\hspace{1cm}\text{or}\hspace{1cm}x=3$$
We do not examine $x=1$ since it is already the point $(1,1)$.
- For $x=3$: $y=2-3=-1$. The intersecting point is $(3,-1)$.
So besides $(1,1)$, the normal also intersects the curve at $(3,-1)$.