f(x, y) = ln(\frac{xy}{3} ). ∇f(x, y) = Df(x, y) = (\frac{1}{x}, \frac{1}{y})

Hessian of f = Hf(x, y) = ( \frac{∂[∇f(x,y)]}{∂x} ; \frac{∂[∇f(x,y)]}{∂y} ) = ( \begin{matrix}\frac{∂^{2}f}{∂^{2}x} & \frac{∂^{2}f}{∂y∂x} \\ \frac{∂^{2}f}{∂y∂x} & \frac{∂^{2}f}{∂^{2}y} \end{matrix} )

Thus Hf(x, y) = ( \begin{matrix}−x^{−2} & 0 \\ 0 & −y^{−2} \end{matrix} )

Let a = (3, 1). Then f(3, 1) = ln(\frac{(3)(1)}{3}) = ln(1) = 0. ∇f(3, 1) = Df(3, 1) = (\frac{1}{3}, 1)

Hf(3, 1) = ( \begin{matrix}−3^{−2} & 0 \\ 0 & −1^{−2} \end{matrix} ) = ( \begin{matrix}−\frac{1}{9} & 0 \\ 0 & −1 \end{matrix} )

First order approximation of f near a = (3, 1): Tangent plane to f(x, y) = ln(x, y) at a = (3, 1) is

p_{1}(x, y) = f(a) + Df(a)(x − a)

p_{1}(x, y) = f(3, 1) + Df(3, 1) ( \begin{matrix}x − 3 \\ y − 1 \end{matrix} ) = 0 + (\frac{1}{3}, 1) ( \begin{matrix}x − 3 \\ y − 1 \end{matrix} )

= \frac{1}{3}(x − 3) + y − 1 = \frac{1}{3}x − 1 + y − 1.

Thus p_{1}(x, y) = \frac{1}{3}x + y − 2.

Second order approximation of f near a = (3, 1): p_{2}(x, y) = f(a) + Df(a)(x − a) + \frac{1}{2}(x − a)^{T} Hf(a)(x − a)

p_{2}(x, y) = \frac{1}{3}x + y − 2 + \frac{1}{2}(x − 3, y − 1) ( \begin{matrix}−\frac{1}{9} & 0 \\ 0 & −1 \end{matrix} ) ( \begin{matrix}x − 3 \\ y − 1 \end{matrix} )

p_{2}(x, y) = \frac{1}{3}x + y − 2 + \frac{1}{2}(x − 3, y − 1) ( \begin{matrix}−\frac{1}{9}(x − 3) \\ −(y − 1) \end{matrix} )

p_{2}(x, y) = \frac{1}{3}x + y − 2 − \frac{1}{18}(x − 3)^{2} − \frac{1}{2}(y − 1)^{2}