find the Critical Point?

f(x, y) = y^2 + xy - x^2 - 5y + 2

Partial Derivative:

fx(x, y) = y - 2x = 0

fy(x, y) = 2y + x - 5 = 0

Solving the simultaneous equation gives x = 1, y = 2

Second Derivative Test

fxx(x, y) = -2x ==> fxx(1, 2) = -2 < 0

fyy(x, y) = 2 ==> fyy(1, 2) = 2

fxy(x, y) = 1 ==> fxy(1, 2) = 1

D(1, 2) = fxx(1, 2) * fyy(1, 2) - [fxy(1, 2)]² = (-2) * 2 - 1² = -5 < 0

Thus f(1, 2) is a critical point and it is a saddle point.

It is the solution of df/dx=0 and df/dy=0, i.e.

y - 2x = 0, and

2y + x - 5 = 0