Proving Vector Subspaces Question 1:
The set $\mathbb R^3$ of all column vectors of length three, with real entries, is a vector space. Is the subset $$B=\{xyz \in \mathbb R^3 \mid xy+yz= 0\}$$ a subspace of $\mathbb R^3$? 
Justify your answer.
Attempted answer:
$(0)$: let $0$ vector be in set $B$, then $0\cdot0 + 0\cdot0 = 0$, hence $B$ is non-empty.
$(1)$ A1: Let $u = \begin{bmatrix}0&0&1\end{bmatrix}$ and $V = \begin{bmatrix}0&1&0\end{bmatrix}$, which are both in $B$, but $u +v = \begin{bmatrix}0&1&1\end{bmatrix}$ is not in $B$, since $0\cdot1 +1\cdot1 \neq 0$. Hence not closed under multiplication.
$(2)$ S1: I know its irrelevant since its not closed under A1...but is this correct--> let $u = \begin{bmatrix}a&b&c\end{bmatrix}$, then $k\cdot u = \begin{bmatrix}ka&kb&kc\end{bmatrix}\implies (ka)(kb) + (kb)(kc) = 0 \implies k^2(ab) + k^2(bc) = 0$, hence must clearly be in set $B$?
Question 2
Show that the set of all twice differentiable functions $f:\mathbb R\to \mathbb R$ satisfying the differential equation $\sin(x)f''(x) + x^2f(x) = 0$ is a vector space with respect to the usual operations of addition of functions and multiplication by scalars. Note that $f''$ is the second derivative of $f$.
Attempted answer: The zero vector in $F$ space has the zero function $0(x) = 0 \in \mathbb R$ and we have $0''(x) = 0'(x) = 0$. Assume that the set of solutions for the equation is a subset of $F$ space, hence it contains the zero function and so $0(x) = \sin(x)\cdot 0''(x) + x^2\cdot0(x) = 0+0 = 0$ so set of solutions is in subspace.

So for question 2:
A1: $\sin(x)(f+g)'' + x^2 (f+g)(x)= 0 \implies 
\sin(x) [f''(x) + g''(x)] + x^2\cdot f(x)+x^2\cdot g(x) = 0$ $ \implies \sin(x)\cdot f''(x) + \sin(x)\cdot g''(x) + x^2\cdot f(x) +x^2\cdot g(x)$ so clearly $f +g \in F$.  
S1: $\alpha \cdot (\sin(x)\cdot f''(x) + x^2\cdot f(x)) \implies \alpha \sin(x)\cdot f''(x) + \alpha \cdot x^2\cdot f(x)$.....stuck here, how do i tell if its in the subset or not ??? really confused in this regard.
