Find where the curve 2x^2 + xy + y^2 = 14 has stationary points

d/dx (xy) = x dy/dx + y 

d/dx (y^2) = 2y dy/dx [This is from the chain rule]

So, d/dx (2x^2 + xy + y^2 = 14) 

=> 4x + x dy/dx + y + 2y dy/dx = 0

set dy/dx = 0 as stationary point has gradient 0

Obtains 4x+y=0

y=-4x

Sub this back into our original equation

14x^2 = 14

x^2 = 1

This is only satisfied by +1 and -1

When x=1 y=-4, when x=-1 y=4

So stationary points are (1,-4) and (-1,4)