Find coordinates where tangent is parallel to $x$-axis given formula for gradient/derivative 
Find the coordinates of the points where the line tangent to the curve $$x^2-2xy+2y^2=4$$ is parallel to the $x$-axis, given that $$\frac{dy}{dx}=\frac{y-x}{2y-x}$$

By letting $dy/dx = 0$ I get $y=x$ which is no help... what do I do?
Thanks
 A: All points on the curve obey the equation $x^2 - 2xy + 2y^2 = 4$. Having $y=x$ when $\frac{dy}{dx} = 0$ reduces this to $x^2 - 2x^2 + 2x^2 = 4$ or $x^2= 4$ or $x=2$ or $x=-2$, which you can find the possible $y$-coordinates for by using the equation again (substitute $x=2$ and solve for $y$, then the same for $x=-2$) so you do get your points that way. Don't forget the base equation after you compute differentials!
A: You are definitely on the right track!
Yes, you should get $y=x$. Just plug it into the original equation and solve (because now you have two conditions that the point should satisfy: the equation and its derivative):
$x^2-2x^2+2x^2=4$
$x=\pm 2$ --> $y=\pm2$
Final points: $(-2,-2), (2,2)$
A: $$x^2-2xy+2y^2=4$$
$$\frac{d}{dx}(x^2-2xy+2y^2)=\frac{d}{dy}(4)$$
$$2x-2(y+x\frac{dy}{dx})+4y\frac{dy}{dx}=0$$
$$2x-2y-2x\frac{dy}{dx}+4y\frac{dy}{dx}=0$$
$$\frac{dy}{dx}=\frac{y-x}{2y-x}$$
Now equate $\frac{dy}{dx}=0$
$$\frac{y-x}{2y-x}=0$$
$$x=y,x=2y$$
Solve by substitution, I substituted $y=x$ in $x^2-2xy+2y^2=4$ and got
$$x^2-2x^2+2x^2=4$$
$$x=-2,2$$
Now substitute each $x$ value in $x^2-2xy+2y^2=4$ and we get$$y=-2,2$$
The points are $(-2,-2),(2,2)$
