Perplexed

Consider equation 1:

$\begin{aligned} \color{#3D99F6} \frac {\frac {4xy}{y^2-x^2}}{\frac 1{y^2-x^2}+\frac 1{y^2+2xy+x^2}} & = 2 & \small \color{#3D99F6} \text{Multiplied up and down by }(y^2-x^2)(y^2+2xy+x^2) \\ \color{#3D99F6} \frac {4xy(y^2+2xy+x^2)}{2y^2+2xy} & = 2 \\ \frac {2x(y+x)^2}{y+x} & = 2 \\ x^2+xy & = 1\end{aligned}$

$\begin{aligned} \implies x^2 & = 1 - xy & ... (1a) \end{aligned}$

Now consider equation 2:

$\begin{aligned} 3x^2 + y^2 + 2x-2y - 1 & = 0 \\ x^2 +2{\color{#3D99F6}x^2} + y^2 + 2x-2y - 1 & = 0 & \small \color{#3D99F6} (1a): x^2 = 1-xy \\ x^2 +2{\color{#3D99F6}(1-xy)} + y^2 + 2x-2y - 1 & = 0 \\ x^2 - 2xy + y^2 + 2x-2y + 1 & = 0 \\ (x-y)^2 + 2(x-y) + 1 & = 0 & \small \color{#3D99F6} \text{It is a quadratic equation of }(x-y). \\ ((x-y)+1)^2 & = 0 \end{aligned}$

$\begin{aligned} \implies y & = x + 1 & ...(2a) \end{aligned}$

Substituting (2a) in (1a):

$\begin{aligned} x^2 & = 1 - x(x+1) \\ 2x^2 + x -1 & = 0 \\ (2x-1)(x+1) & = 0 \end{aligned}$

$\implies \begin{cases} x = \frac 12 & \text{(2a): } y = \dfrac 12 + 1 = \dfrac 32 & \implies 3y(x+y) - 1 = \boxed{8} \\ x = -1 & \text{(2a): } y = -1 + 1 = 0 & \color{#D61F06} \small \text{But (-1,0) does not satisfy equation (1).} \end{cases}$