Well, we will use the differential of the function $f(x,y) = \sqrt{x^2-y^2}$ to get an estimate of the variation $df$, then we can calculate the new value of $f$ ($f_n$).

$$\begin{gather*} \because f(x,y) = \sqrt{x^2 - y^2} \\ \text{for $\sqrt{(4.98)^2 - (3.03)^2 }$} \\ \therefore x = 5 \:\:,\:\: dx = -0.02 \:\:,\:\: y = 3 \:\:,\:\: dy = 0.03 \\ \therefore f(5,3) = \sqrt{5^2 - 3^2} = \sqrt{16} = \mathbf{4} \\ \because df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy\\ \therefore df = \frac{x dx}{(x^2-y^2)^{\frac{1}{2}} } - \frac{y dy}{(x^2-y^2)^{\frac{1}{2}} } \\ \therefore df = \frac{5\times -0.02 - 3\times 0.03}{4} = \frac{-0.10 - 0.09}{4} \\ \therefore df = \frac{-0.19}{4} = \mathbf{-0.0475} \\ \because f_n = f + df \\ \boxed{ \therefore f_n = 4- 0.0475 = 3.9525 \approx \mathbf{3.95} } \end{gather*}$$