Answer
$(3,-1)$
Work Step by Step
Step 1. Given the equation $x^2+2xy-3y^2=0$, differentiate both sides with respect to $x$: $2x+2y+2xy'-6yy'=0$, which gives $y'=\frac{x+y}{3y-x}$,
Step 2. The slope of the tangent line at the given point $(1,1)$ is $m=y'=\frac{1+1}{3-1}=1$
Step 3. The slope of the line normal to the tangent line is $n=-\frac{1}{m}=-1$, and the line equation is $y-1=-(x-1)$ or $y=-x+2$
Step 4. Use the above equation with the original: $x^2+2x(-x+2)-3(-x+2)^2=0$ or $x^2-2x^2+4x-3x^2+12x-12 =0$ or $x^2-4x+3 =0$. Thus we have $x=1, 3$. Using $x=3$, we have $y=-3+2=-1$, which gives the point $(3,-1)$
