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?
 A: Subbing $x$ into the original equation isn't wrong, it's just not enough.  This is because $x^2 - 2xy + 2y^2 = 4$ does not express $y$ as a function of $x$.
With implicit functions, it's not enough to just plug the $x$-coordinates back into the original function, because that doesn't tell you which $x$- and $y$-coordinates correspond to points with the desired slope.  This is why you also need to evaluate $dy/dx$ at the points you find.  This is not an issue with explicit functions because each $x$-coordinate corresponds to at most one $y$-coordinate.
Another way of looking at it:  For this problem, you've found that $dy/dx = 0$ implies $x = \pm 2$.  This is correct.  But this does not mean that $x= \pm 2$ implies $dy/dx = 0$.  And again, this is not an issue with explicit functions because for explicit functions it isn't possible for one $x$-coordinate to correspond to more than one point on the graph (and therefore potentially more than one derivative or slope).
A: 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. 
