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

We have the implicit curve given by $$x^2y^2-2x=4(1-y).$$

First we rearrange everything to the left so that the expression is equal to zero:

$$x^2y^2-2x-4(1-y)=0.$$

Simplifying the bracket, we get

$$x^2y^2-2x-4+4y=0.$$

Let us denote $$F(x,y)=x^2y^2-2x-4+4y.$$ The tangent’s slope at any point is obtained from the total derivative $$\dfrac{dF}{dx}=0,$$ that is the rule

$$\frac{\partial F}{\partial x}+\frac{\partial F}{\partial y}\,\frac{dy}{dx}=0.$$

We differentiate term by term.

The partial derivative with respect to $$x$$ is $$\frac{\partial F}{\partial x}=2xy^2-2.$$

The partial derivative with respect to $$y$$ is $$\frac{\partial F}{\partial y}=2x^2y+4.$$

Putting these into the formula,

$$\bigl(2xy^2-2\bigr)+\bigl(2x^2y+4\bigr)\frac{dy}{dx}=0.$$

So the derivative (slope) is

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

We now evaluate this slope at the given point $$(2,-2).$$

Numerator:

$$2xy^2-2=2\cdot2\cdot(-2)^2-2=2\cdot2\cdot4-2=16-2=14.$$

Denominator:

$$2x^2y+4=2\cdot2^2\cdot(-2)+4=2\cdot4\cdot(-2)+4=-16+4=-12.$$

Thus

$$\frac{dy}{dx}=-\frac{14}{-12}=+\frac{7}{6}.$$

Therefore the slope of the tangent at $$(2,-2)$$ is $$\displaystyle m=\frac{7}{6}.$$

The point-slope form of a straight line is $$y-y_1=m(x-x_1).$$ Using $$(x_1,y_1)=(2,-2)$$ and $$m=\frac{7}{6},$$ we write

$$y-(-2)=\frac{7}{6}(x-2).$$

That is

$$y+2=\frac{7}{6}(x-2).$$

To clear the fraction, multiply by $$6$$:

$$6(y+2)=7(x-2).$$

So

$$6y+12=7x-14.$$

Bringing all constants to one side,

$$6y=7x-26,$$ and finally

$$y=\frac{7}{6}x-\frac{26}{6}=\frac{7}{6}x-\frac{13}{3}.$$

Now we test each option by substituting its coordinates into this line equation.

Option A: $$(-2,-7)$$ Left side: $$y=-7.$$ Right side: $$\frac{7}{6}(-2)-\frac{13}{3}=-\frac{14}{6}-\frac{13}{3}=-\frac{7}{3}-\frac{13}{3}=-\frac{20}{3}\;(\approx-6.67).$$ Since $$-7\neq-\dfrac{20}{3},$$ the point $$(-2,-7)$$ is not on the tangent.

Option B: $$(8,5)$$ Right side: $$\frac{7}{6}(8)-\frac{13}{3}=\frac{56}{6}-\frac{26}{6}=\frac{30}{6}=5.$$ The left side is also $$5,$$ so $$(8,5)$$ lies on the tangent.

Option C: $$(-4,-9)$$ Right side: $$\frac{7}{6}(-4)-\frac{13}{3}=-\frac{28}{6}-\frac{26}{6}=-\frac{54}{6}=-9,$$ which equals the given $$y,$$ so $$(-4,-9)$$ also lies on the tangent.

Option D: $$\left(4,\dfrac{1}{3}\right)$$ Right side: $$\frac{7}{6}(4)-\frac{13}{3}=\frac{28}{6}-\frac{26}{6}=\frac{2}{6}=\frac{1}{3},$$ matching the given $$y,$$ so this point lies on the tangent as well.

Thus the only point through which the tangent does not pass is $$(-2,-7).$$

Hence, the correct answer is Option A.