Given $x^2+2xy-3y^2=0$

$\left(x+3y\right)\left(x-y\right)=0$

Pair of straight lines passing through the origin.

$\because x+3y=0$  or  $x-y=0$

Normal exists at $\left(1, 1\right)$ which is on $x-y=0$

$\Rightarrow$  The slope of normal at $\left(1, 1\right)=-1$ 

$\therefore$  Equation of normal will be

$y-1=-\left(x-1\right)$    $\left[y-y_1=m\left(x-x_1\right)\right]$

$y-1=-x+1$

$x+y=2$

Now, find the point of intersection with $x+3y=0$. 

   $\begin{array}{c}x+y=2\\ x+3y=0\end{array}$
  ----------------
   $-2y=2$   $\Rightarrow y=-1, x=3$

$\because \left(3, -1\right)$ lies in the fourth quadrant.