# Find all critical points, use second partial test to classify as relative Max, Min or saddle point f(x,Y)= xy- 1/y -1/x

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You need to find the first order partial derivatives such that:

`f_x(x,y) = y + 1/x^2` (y is kept constant and the expression is differentiated with respect to x)

`f_y(x,y) = x + 1/y^2` (x is kept constant and the expression is differentiated with respect to y)

You need to find the critical points of the function, hence, you need to solve the equations:

`{(f_x(x y) = 0) , (f_y(x y) = 0):}`

`y + 1/x^2 = 0 =gt x^2*y = -1 =gt y = -1/x^2`

`x + 1/y^2 = 0 =gt1/x^3 = -1 =gt x^3 +1 = 0`

`(x+1)(x^2 - x + 1) = 0 =gt x = -1 =gt y = -1`

Hence, the function has a critical point at `(-1,-1).`

You need to classify the critical point, hence, you need to evaluate D such that:

`D = f_(x x)(-1,-1)*f_(y y)(-1,-1) - f^2_(x y)(-1, -1)`

`f_(x x) = -2x/x^4 => f_(x x) = -2/x^3 => f_(x x)(-1,-1) = 2`

`f_(y y) = -2/y^3 => f_(y y)(-1,-1) = 2`

`f_(x y) = 1`

`D = 2*2 - 1 => D = 3`

**Hence, since `D = 3 > 0` and `f_(x x)(-1,-1) = 2 > 0,` the function has a relative minimum at (-1,-1).**

**Sources:**