Answer
See below
Work Step by Step
We can write set $S$ as $S=\{(x,y) \in R^2: x^2-y^2=0\}$
We have $(1,1) \in S$ and $(1,-1) \in S$, obtain $(1,1)+(1,-1)=(2,0) \notin S$.
Hence $S$ is not closed under addition in $R^2$.
$S$ is not a subspace of $R^2$
