The point of intersection of the tangents drawn to the curve $x^2 y=1-y$ at the points where it is met by the curve $x y=1-y$, is given by

(0, 1)

Solving the two equations, we get

$x^2y=xy \Rightarrow xy(x-1) = 0 \Rightarrow x = 0, y = 0, x= 1$

Since y = 0 does not satisfy the two equations. So, we neglect it. Putting x = 0 in the either equation, we get x = 1

Now, putting x = 1 in one of the two equations we obtain y = 1/2. Thus, the two curves intersect at (0, 1) and (1, 1/2).

Now,

$x^2 y=1-y \Rightarrow x^2 \frac{d y}{d x}+2 x y =-\frac{d y}{d x} \Rightarrow \frac{d y}{d x}=-\frac{2 x y}{x^2+1}$

$\Rightarrow \left(\frac{d y}{d x}\right)_{(0,1)}=0$ and $\left(\frac{d y}{d x}\right)_{(1,1 / 2)}=-\frac{1}{2} $

The equations of the required tangents are

$y-1=0(x-0)$ and $y-\frac{1}{2}=\frac{-1}{2}(x-1)$

$\Rightarrow y=1$ and $x+2 y-2=0$

These two tangents intersect at (0, 1).