The tangent at the point (2,  -2) to the curve $x^2 y^2-2x=4(1-y)$ does not passes through the point

Curve: \(x^2y^2-2x=4(1-y)\). Point: \((2,-2)\).

Differentiate implicitly:
\(\dfrac{d}{dx}(x^2y^2)-2=\dfrac{d}{dx}\big(4(1-y)\big)\)
\(2xy^2+2x^2y\,y'-2=-4y'\)
\(\Rightarrow y'\,(2x^2y+4)=2-2xy^2\)
\(\Rightarrow y'=\dfrac{1-xy^2}{x^2y+2}\).

Slope at \((2,-2)\):
\(xy^2=2\cdot4=8\), \(x^2y+2=4\cdot(-2)+2=-6\).
\(m=y'=\dfrac{1-8}{-6}=\dfrac{-7}{-6}=\dfrac{7}{6}\).

Tangent line at \((2,-2)\):
\(y+2=\dfrac{7}{6}(x-2)\ \Rightarrow\ 6y=7x-26\ \Rightarrow\ \boxed{\,y=\tfrac{7}{6}x-\tfrac{13}{3}\,}\).

How to decide “does not pass through”: A point \((x_0,y_0)\) lies on the tangent iff \(6y_0=7x_0-26\). If this fails, the tangent does not pass through that point.

Checks (examples): On the line: \((0,-\tfrac{13}{3})\), \((\tfrac{26}{7},0)\). Any point not satisfying \(6y=7x-26\) is not on the tangent.