Given curve $x^2{y}^2-2x=4-4y$

Differentiate both sides w.r.t  $x$,

$\Rightarrow$ $2x{y}^2+2y.x^2.\frac{dy}{dx}-2=-4.\frac{dy}{dx}$

$\Rightarrow$ $\frac{dy}{dx}\left(2y.x^2+4\right)=2-2x.y^2$

$\Rightarrow$ ${\left.\frac{dy}{dx}\right|}_{\left(2,-2\right)}=\frac{2-\left(2\times 2\times 4\right)}{2\left(-2\right)\times 4+4}=\frac{14}{12}=\frac{7}{6}$

Equation of tangent, $\left(y+2\right)=\frac{7}{6}\left(x-2\right)$

$\Rightarrow 7x-6y=26$, which doesn't pass through $\left(-2-7\right)$.