Answer
$$-\sqrt{\dfrac{2}{3}}$$
Work Step by Step
$$ Surface \space Area =\int_0^1 \int_0^1 [(u+v)(u-v)-v^2] \times \sqrt 6 \space du \space dv \\=\sqrt 6 \times \int_0^1 u^2-2v^2 \space du \space dv \\=\sqrt 6 \times \int_0^1 [\dfrac{u^3}{3}-2uv^2]_0^1 \space
dv \\ = \sqrt 6[\dfrac{v}{3}-\dfrac{2v^3}{3}]_0^1 \\=-\sqrt{\dfrac{2}{3}}$$
