Verified answer

Differentiate both sides with respect to x:

(2x * y + x^2 * 1 dy/dx) - (1 * y^2 + x * 2y dy/dx) = 0.

Solve for dy/dx:

(2xy - y^2) + (x^2 - 2xy) dy/dx = 0

==> dy/dx = (y^2 - 2xy)/().

We have vertical tangents when dy/dx is infinite (that is, set the denominator equal to 0):

x^2 - 2xy = 0

==> x(x - 2y) = 0

==> x = 0 or x = 2y.

If x = 0, then 0^2 * y - 0 * y^2 = 2 has no solution.

If x = 2y, then (2y)^2 y - (2y)y^2 = 2

==> y = 1.

So, the only point is (2, 1).

I hope this helps!