Find the coordinates of the points on the curve X^2 -2xy+2y^2=4 at which the tangent is parallel to x-axis.
You don't really have to find the derivative explicitely. Going back to your original equation, $\displaystyle x^2 -2xy+2y^2=4$, implicit differentiation gives $\displaystyle 2x- 2y- 2xy'z+ 4yy'= 0$. Now, because you are looking for points where y'= 0, put y'= 0 into that equation: $\displaystyle 2x- 2y= 0$. That tells you that y= x. Replace y with x in the original equation, $\displaystyle x^2 -2xy+2y^2=4$ to get an equation in x only and solve that for x.