Implicit differentiation : two or four tangents? 
My Question is: The equation of a curve is $$ x ^ 2 - 2 x y + 2 y ^ 2 = 4 $$ Find the coordinates of each point on the curve at which the tangent is parallel to the $ x $-axis.

So I established that $$ \frac { \mathop { \sf d } y } { \mathop { \sf d } x } = 0 \implies y = x $$ Therefore $ x = \pm 2 $ by subbing $ x = y $ in to the original equation. The problem I'm having is that if I sub $ x = \pm 2 $ in to the original equation I get $ 4 $ coordinates. But when I sub it in to $ \frac { \mathop { \sf d } y } { \mathop { \sf d } x } $ I get $ 2 $ coordinates?
The correct answer is $ ( 2 , 2 ) $ and $ ( 2 , - 2 ) $ which is from subbing $ x $ in to $ \frac { \mathop { \sf d } y } { \mathop { \sf d } x } = 0 $.
Why isn't subbing $ x $ in to the original equation correct?

The Points
See that 
$$x^2 - 2xy + 2y^2 = 4 \implies (x - y)^2 = 2^2 - y^2 = (2-y)(2+y)$$
So we have that 

$$(x - y)^2 = (2-y)(2+y)$$

If we differentiate the equation we get that, using the notation $y' := \mathrm dy/\mathrm d x$ 
$$2x - 2y - 2xy' + 4yy' = 0 \implies y' = \frac{y - x}{2y - x}$$
Setting $y' = 0$ implies that $y = x$ and if you use this result in the above equation you'll find
$$(2-y)(2+y) = 0$$
So you must have that $y = 2$ or $y = -2$ then you get the results $(2,2)$ and $(-2,-2)$ and as @tilper pointed out
$$y'(2,2) = y'(-2,-2) = 0$$
note that $y'(2,-2) = 4/6 \neq 0$ so in this point you do not get the tangent parallel. 
These are the only points?
Yes. To find horizontal tangents we have to find points where $y'= 0$. So we have in this case that the only points were the derivative is equal to zero are the points where
$$y = x \hspace{0.3cm} \text{for points }(x,y)\in \text{Dom}(f)$$
where $f(x,y) = x^2 - 2xy + 2y^2$. We are looking for the curve $f(x,y)=4$ then the points that have $x = y$ will be such that pointed out above. These are the only solutions to this curve so they are the only points. By the definition the tangent is horizontal when the point is stationary and they occur only when the derivative is zero. It is important that the derivative $y'$ is well defined, e.g., in the points $2y = x$ the derivative is not defined.
See this question where the answer given is such that, to demonstrate that there is no horizontal tangent,  the author proves that $y'\neq 0$ for all points in the domain. 
