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?

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).
