Answer
$W$ is not a subspace of $R^2$.
Work Step by Step
Let $W$ be a subset of $V$ such that
$$W=\left\{(x, y) : y=a x^{2}\right\}, \quad V=R^{2}.$$
Assume that $u=(x,ax^2), v=(y,ay^2)\in W$ and $c\in R$. Now, we have
(a) $W$ contains the zero vector $(0,0)$.
(b) $u+v=(x,ax^2)+(y,ay^2)=(x+y,ax^2+ay^2)=(x+y,a(x^2+y^2))$.
Since $x^2+y^2\neq (x+y)^2$, then $u+v\not \in W$. Hence, $W$ is not a subspace of $R^2$.
