Find $\frac{dy}{dx}$ equal to zero in terms of x and y

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The original equation: $x^2 - xy + y^2 = 3$

After Implicit Differentiation: $\dfrac{\mathrm dy}{\mathrm dx} = (-2x+y)(-x+2y)$

If I want to find $x$ and $y$ when $ \frac{\mathrm dy}{\mathrm dx} = 0$ how would I do that?

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Solve -2x + y = 0 and -x + 2y = 0. The values where the two equations equal zero is where dy/dx will equal zero as well. Although, as mentioned above, one or the other will be zero, and since anything multiplied by zero is zero, the whole equation will equal zero.

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$dy/dx$ is the slope at all the tangents. When $dy/dx = 0 $ so is the slope,

so when $(-2x + y )( -x+2y) = 0 $ either $-2x +y= 0$ or $-x+2y=0$ (as the product of two non-zero numbers cannot be zero). You basically have a very simple simultaneous equation to solve for $x$ and $y$.