Use a tangent plane to approximate the value of the following function at the point $(-3.9,7.9)$ : \[ f(x, y)=\arctan \left(x y^{2}+257\right) \] Present you answer.accurate to four decimal places.

Step 1 ：Define the function \(f(x, y)=\arctan \left(x y^{2}+257\right)\).

Step 2 ：Find the partial derivatives of the function with respect to \(x\) and \(y\), which are \(f_x = \frac{y^2}{(x*y^2 + 257)^2 + 1}\) and \(f_y = \frac{2*x*y}{(x*y^2 + 257)^2 + 1}\) respectively.

Step 3 ：Evaluate the partial derivatives at the point \((-3.9,7.9)\), obtaining \(f_x(-3.9,7.9) = 0.335560724955477\) and \(f_y(-3.9,7.9) = -0.331313120842116\).

Step 4 ：Calculate the value of the function at the point \((-3.9,7.9)\), which is \(f(-3.9,7.9) = 1.49740437800057\).

Step 5 ：Substitute these values into the equation of the tangent plane, \(z = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)\), to get \(z = 1.49740437800057\).

Step 6 ：Round the result to four decimal places to get the final answer, \(\boxed{1.4974}\).