Answer
$$A(S) \leq \pi \sqrt 3 R^2$$
Work Step by Step
We are given that $|f_x| \leq 1$ and $(f_x)^2 \leq 1$ and $|f_x| \leq 1 \implies (f_y)^2 \leq 1$
and $A(S)=\iint_{D} \sqrt {1+(f_x )^2+(f_y)^2 } dA $; $\iint_{D} dA$ is the area of the region $D$
So, , $\sqrt {1+(f_x )^2+(f_y)^2 } \leq \sqrt {1+1+1} \\ \sqrt {1+(f_x )^2+(f_y)^2 } \leq \sqrt 3$
Now, $$A(S) \leq \sqrt 3 \iint_{D} dA$$
Since, $x^2+y^2 \leq R^2$ and the area of the region $D$ is $\pi R^2$
This implies that $ \iint_{D} dA= \pi R^2 $
Therefore, $A(S) \leq \pi \sqrt 3 R^2$
