We need to differentiate this with respect to x to see if we can find an expression for the derivative of y at various points. That will be the slope of the tangent to the curve. Then we want to see where that derivative might be infinite -- i.e., where the tangent is vertical.
It's not written as a function, but it can still be differentiated using the chain rule:
x2 + xy + y2 = 3
(2x) + (x dy/dx + y dx/dx) + (2y dy/dx) = 0
(I used parentheses to show the differentiation of each term in the original equation.)
2x + x dy/dx + y + 2y dy/dx = 0
2x + y = -x dy/dx - 2y dy/dx
2x + y = dy/dx (-x -2y)
-(2x + y)/(x + 2y) = dy/dx
We have the derivative of y, but it's defined partly in terms of y itself. That's OK. Let's go on...
So where would the slope be infinite? That would happen when x + 2y = 0, or y = -x/2
Let's plug that in for y in the original equation to find points where that's the case.
x2 + xy + y2 = 3
x2 + x(-x/2) + (-x/2)2 = 3
x2 - x2/2 + x2/4 = 3
3x2 / 4 = 3
x2 = 4
x = ±2
So we have two x values where the tangent might be vertical. Let's plug them into the equation and see what the y values are. First x = 2...
x2 + xy + y2 = 3
4 + 2y + y2 = 3
y2 + 2y + 1 = 0
(y + 1)2 = 0
y = -1
So at the point (2, -1) the tangent is vertical.
Now try x = -2...
x2 + xy + y2 = 3
4 - 2y + y2 = 3
y2 - 2y + 1 =0
(y - 1)2 = 0
y = 1
So at the point (-2, 1) the tangent is vertical.
