To estimate the value of Z at x1=1.98 and y1=3.06 using the given data points x0=2, y0=3, we can first calculate the average rate of change of Z with respect to x:

ΔZ/Δx = (Z1 - Z0) / (x1 - x0)

Next, we can use the chain rule to express the derivative of Z with respect to x in terms of the partial derivatives of Z with respect to x and y:

dZ/dx = (∂Z/∂x) + (∂Z/∂y) * (dy/dx)

Since we have the values of x0, x1, y0, y1, we can calculate the values of ΔZ, ∂Z/∂x, ∂Z/∂y, and dy/dx to estimate Z at x1=1.98 and y1=3.06. Let's calculate these values:

Given:
x0 = 2, x1 = 1.98
y0 = 3, y1 = 3.06

Δx = x1 - x0 = 1.98 - 2 = -0.02
Δy = y1 - y0 = 3.06 - 3 = 0.06

ΔZ = ln(y1^2 - 2x1) - ln(y0^2 - 2x0)
ΔZ = ln(3.06^2 - 2(1.98)) - ln(3^2 - 2(2))

Calculating ΔZ, we get:
ΔZ = ln(9.3636) - ln(5)
ΔZ ≈ ln(9.3636) - 1.6094
ΔZ ≈ 2.235 - 1.6094
ΔZ ≈ 0.6256

∂Z/∂x = (∂Z/∂x)y + (∂Z/∂y)(dy/dx)
∂Z/∂x = (∂/∂x)(ln(y^2 - 2x)) + (∂/∂y)(ln(y^2 - 2x))(dy/dx)
∂Z/∂x = -2/(y1^2 - 2x1) + 0
∂Z/∂x = -2/(3.06^2 - 2*1.98)
∂Z/∂x = -2/(9.3636 - 3.96)
∂Z/∂x = -2/(5.4036)
∂Z/∂x ≈ -0.3701

∂Z/∂y = (∂/∂y)(ln(y^2 - 2x))
∂Z/∂y = 2y/(y^2 - 2x)
∂Z/∂y = 23.06/(3.06^2 - 21.98)
∂Z/∂y = 6.12/(9.3636 - 3.96)
∂Z/∂y = 6.12/(5.4036)
∂Z/∂y ≈ 1.1323

dy/dx = Δy/Δx
dy/dx = 0.06 / -0.02
dy/dx = -3

Now, we can use these values to estimate Z at x1=1.98 and y1=3.06:
Z1 = Z0 + ΔZ
Z1 = ln(y0^2 - 2x0) + ΔZ
Z1 = ln(3^2 - 2*2) + 0.6256
Z1 = ln(9-4) + 0.6256
Z1 = ln(5) + 0.6256
Z1 ≈ 1.6094 + 0.6256
Z1 ≈ 2.235

Therefore, the estimated value of Z at x1=1.98 and y1=3.06 is approximately 2.235.