Answer
$0$
Work Step by Step
Stoke's Theorem: Let us consider that $C$ be the closed curve which encloses surface $S$ for a vector field $F$. This also states that the line integral and the surface integral must be equal. That is,
$\oint_C F \ dx=\iint_S (\nabla \times F) \cdot n \ dS$
Here, $\nabla \times F=\begin{vmatrix} i&j&k \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ x^2-y^2 & z^2-x^2 & y^2-z^2 \end{vmatrix}=2 (y-z,0, -x+y)$ and $n=(-z_x, -z_y,1) =(0,0,1)$
Thus, the surface integral can be computed as:
$\iint_S (\nabla \times F) \cdot n \ dS=\iint_S 2 (y-z,0, -x+y) \cdot (0,0,1) \ dS\\=2 \iint_S -x+y dS\\=2 \int_{-1}^1 \int_{-1}^{1} -x+y \ dx \ dy \\= \int_{-1}^1 [-x^2+2xy]_{-1}^{1} \\=4 \int_{-1}^1 y \ dy\\=2 \times [y^2]_{-1}^1 \\=0$
